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Hess's law state function diagram explanation

General Chemistry

Thermochemistry.

In this set of practice questions, we will go over the main types of questions on calorimetry including the heat capacity, the heat of reaction, finding the final temperature of a mixture, constant pressure calorimetry, and constant-volume calorimetry.

A few important concepts and formulas you will need to solve these questions. 

Heat capacity and specific heat are correlated by the following formula: 

solving calorimetry problems

When solving a problem related to heat capacity and heat transfer, remember that most of the time, it is assumed the heat is not lost, and it only flows from the object with a higher temperature to the colder one:

solving calorimetry problems

The heat transfer to/from the calorimeter is determined by its temperature change and the heat capacity:

q cal  = C cal  x Δ T

The heat capacity of the calorimeter is determined experimentally and is already known when measuring the heat/enthalpy of a reaction.

Remember, the enthalpy change  ( Δ H)  is equal to the heat  when the  pressure is constant :

solving calorimetry problems

On the other hand, a bomb calorimeter  is an equipment that measures  Δ E  for combustion reactions . This can be seen in the equation of internal energy change:

Δ E = q + w

w =  P Δ V  and Δ V  is zero when the volume is constant. So, for a constant volume,  Δ E  =  q , and therefore, the  bomb calorimeter measures the energy change  of a chemical reaction.

solving calorimetry problems

The experiment is carried out in an insulated sealed vessel called a  bomb  which is designed to withstand high pressures. The bomb is placed in a water container and by measuring its temperature change caused by the reaction, we determine the heat of the calorimeter.

solving calorimetry problems

The links to the corresponding topics are given below:

  • Energy Related to Heat and Work
  • Endothermic and Exothermic Processes
  • Heat Capacity and Specific Heat
  • Heat Capacity Practice Problems
  • What is Enthalpy
  • Constant-Pressure Calorimetry
  • Bomb calorimeter – Constant Volume Calorimetry
  • Stoichiometry and Enthalpy of Chemical Reactions
  • Hess’s Law and Enthalpy of Reaction
  • Hess’s Law Practice Problems
  • Standard Enthalpies of Formation
  • Enthalpy of Reaction from Enthalpies of Formation

How much heat in kJ is required to warm 1.50 L of water from 25.0 o C to 100.0 °C? (Assume a density of 1.0 g/mL for the water.)

What is the final temperature when a 40 g sample of water at 90 °C is mixed with a 60 g sample at 25 °C?

How much heat does it take to increase the temperature of a 540.6-g sample of Fe from 20.0 °C to 84.3 °C? The specific heat of iron = 0.450 J/g °C.

Calculate the specific heat capacity of a metal if a 17.0 g sample requires 481 J to change the temperature of the metal from 25.0 °C to 67.0 °C?

Calculate the energy of combustion for one mole of butane if burning a 0.367 g sample of butane (C 4 H 10 ) has increased the temperature of a bomb calorimeter by 7.73 °C. The heat capacity of the bomb calorimeter is 2.36 kJ/ °C.

How many joules of energy is required to melt 40.0 g of ice at 0 °C? The heat of fusion (Δ H fus ) for ice is 334.0 J/g.

How many kJ of energy does it take to change 36.0 g of ice at -15.0 °C to water at 0. °C ? The specific heat of ice is 2.10 J/g°C and the heat of fusion (Δ H fus ) for ice is 334.0 J/g. Ignore the significant figures for this problem.

The enthalpy change for the reaction is given below:

2CH 3 OH( l ) + 3O 2 ( g ) → 4H 2 O( l ) + 2CO 2 ( g ) Δ H = -1452.8 kJ

a) What quantity of heat is released for each mole of water formed?

b) What quantity of heat is released for each mole of oxygen reacted?

How much heat will be released if 44.8 g of SO 2  is reacted with an excess of oxygen according to the following chemical equation?

2SO2( g ) + O 2 ( g ) → 2SO 3 ( g ), Δ H ° = –198 kJ

What is Δ H ° for the following reaction

2C 6 H 6 ( l ) + 15O 2 ( g ) → 12CO 2 ( g ) + 6H 2 O( l ), Δ H ° = ? kJ

if the consumption of 27.3 g of benzene (C 6 H 6 ) produces 1144 kJ of heat?

Based on the heat of reaction for the chlorination of methane, how much heat will be released if 233.6 grams of hydrochloric acid are formed?

CH 4 ( g ) + 3Cl 2 ( g ) → CHCl 3 ( l ) + 3HCl( g ), Δ H ° = -334 kJ

Calculate how many kJ of heat-energy will be released when 12.65 g of magnesium carbonate reacts with 650. mL of 0.400  M  hydrochloric acid?

MgCO 3 ( s )   +   2HCl( aq )    →    MgCl 2 ( aq )   +   H 2 O( l )   +   CO 2 ( g ), Δ H °   =   –112   kJ

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  • 1.4 Heat Transfer, Specific Heat, and Calorimetry
  • Introduction
  • 1.1 Temperature and Thermal Equilibrium
  • 1.2 Thermometers and Temperature Scales
  • 1.3 Thermal Expansion
  • 1.5 Phase Changes
  • 1.6 Mechanisms of Heat Transfer
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Molecular Model of an Ideal Gas
  • 2.2 Pressure, Temperature, and RMS Speed
  • 2.3 Heat Capacity and Equipartition of Energy
  • 2.4 Distribution of Molecular Speeds
  • 3.1 Thermodynamic Systems
  • 3.2 Work, Heat, and Internal Energy
  • 3.3 First Law of Thermodynamics
  • 3.4 Thermodynamic Processes
  • 3.5 Heat Capacities of an Ideal Gas
  • 3.6 Adiabatic Processes for an Ideal Gas
  • 4.1 Reversible and Irreversible Processes
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  • 4.4 Statements of the Second Law of Thermodynamics
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  • 4.6 Entropy
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  • 12.1 The Biot-Savart Law
  • 12.2 Magnetic Field Due to a Thin Straight Wire
  • 12.3 Magnetic Force between Two Parallel Currents
  • 12.4 Magnetic Field of a Current Loop
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  • 14.1 Mutual Inductance
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  • 16.1 Maxwell’s Equations and Electromagnetic Waves
  • 16.2 Plane Electromagnetic Waves
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  • 16.4 Momentum and Radiation Pressure
  • 16.5 The Electromagnetic Spectrum
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Explain phenomena involving heat as a form of energy transfer
  • Solve problems involving heat transfer

We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of energy transfer that is caused by a temperature difference, and it can change the temperature of an object. As we learned earlier in this chapter, heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. Heat transfer is fundamental to such everyday activities as home heating and cooking, as well as many industrial processes. It also forms a basis for the topics in the remainder of this chapter.

We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We discuss another way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the study of the relationship of heat and work, which is the basis of engines and refrigerators and the central topic (and origin of the name) of thermodynamics.

Internal Energy and Heat

A thermal system has internal energy , which is the sum of the microscopic energies of the system. This includes thermal energy, which is associated with the mechanical energies of its molecules and which is proportional to the system’s temperature. As we saw earlier in this chapter, if two objects at different temperatures are brought into contact with each other, energy is transferred from the hotter to the colder object until the bodies reach thermal equilibrium (that is, they are at the same temperature). No work is done by either object because no force acts through a distance (as we discussed in Work and Kinetic Energy ). These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure 1.9 shows an example of heat transfer.

The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say “the heat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy flow, whereas temperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature.

Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is the calorie (cal), defined as the energy needed to change the temperature of 1.00 g of water by 1.00 ° C 1.00 ° C —specifically, between 14.5 ° C 14.5 ° C and 15.5 ° C 15.5 ° C , since there is a slight temperature dependence. Also commonly used is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by 1.00 ° C 1.00 ° C . Since mass is most often specified in kilograms, the kilocalorie is convenient. Confusingly, food calories (sometimes called “big calories,” abbreviated Cal) are actually kilocalories, a fact not easily determined from package labeling.

Mechanical Equivalent of Heat

It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of a system. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat — the work needed to produce the same effects as heat transfer . In the units used for these two quantities, the value for this equivalence is

We consider this equation to represent the conversion between two units of energy. (Other numbers that you may see refer to calories defined for temperature ranges other than 14.5 ° C 14.5 ° C to 15.5 ° C 15.5 ° C .)

Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can produce the same effects and measuring the mechanical equivalent of heat. It helped establish the principle of conservation of energy. Gravitational potential energy ( U ) was converted into kinetic energy ( K ), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. Joule’s contributions to thermodynamics were so significant that the SI unit of energy was named after him.

Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although a system has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A well-defined quantity that depends only on the current state of the system, rather than on the history of that system, is known as a state variable . Temperature and internal energy are state variables. To sum up this paragraph, heat and work are not state variables .

Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in the next section, the temperature does not change when a substance changes from one phase to another. An example is the melting of ice, which can be accomplished by adding heat or by doing frictional work, as when an ice cube is rubbed against a rough surface.

Temperature Change and Heat Capacity

We have noted that heat transfer often causes temperature change. Experiments show that with no phase change and no work done on or by the system, the transferred heat is typically directly proportional to the change in temperature and to the mass of the system, to a good approximation. (Below we show how to handle situations where the approximation is not valid.) The constant of proportionality depends on the substance and its phase, which may be gas, liquid, or solid. We omit discussion of the fourth phase, plasma, because although it is the most common phase in the universe, it is rare and short-lived on Earth.

We can understand the experimental facts by noting that the transferred heat is the change in the internal energy, which is the total energy of the molecules. Under typical conditions, the total kinetic energy of the molecules K total K total is a constant fraction of the internal energy (for reasons and with exceptions that we’ll see in the next chapter). The average kinetic energy of a molecule K ave K ave is proportional to the absolute temperature. Therefore, the change in internal energy of a system is typically proportional to the change in temperature and to the number of molecules, N . Mathematically, Δ U ∝ Δ K total = N K ave ∝ N Δ T Δ U ∝ Δ K total = N K ave ∝ N Δ T The dependence on the substance results in large part from the different masses of atoms and molecules. We are considering its heat capacity in terms of its mass, but as we will see in the next chapter, in some cases, heat capacities per molecule are similar for different substances. The dependence on substance and phase also results from differences in the potential energy associated with interactions between atoms and molecules.

Heat Transfer and Temperature Change

A practical approximation for the relationship between heat transfer and temperature change is:

where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and Δ T Δ T is the change in temperature. The symbol c stands for the specific heat (also called “ specific heat capacity ”) and depends on the material and phase. In the SI system, the specific heat is numerically equal to the amount of heat necessary to change the temperature of 1.00 1.00 kg of mass by 1.00 ° C 1.00 ° C . The SI unit for specific heat is J/ ( kg × K ) J/ ( kg × K ) or J/ ( kg × °C ) J/ ( kg × °C ) . (Recall that the temperature change Δ T Δ T is the same in units of kelvin and degrees Celsius.)

Values of specific heat must generally be measured, because there is no simple way to calculate them precisely. Table 1.3 lists representative values of specific heat for various substances. We see from this table that the specific heat of water is five times that of glass and 10 times that of iron, which means that it takes five times as much heat to raise the temperature of water a given amount as for glass, and 10 times as much as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.

The specific heats of gases depend on what is maintained constant during the heating—typically either the volume or the pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the second (in parentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic theory of gases.

In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given in terms of an infinitesimal change in temperature. To do this, we note that c = 1 m Δ Q Δ T c = 1 m Δ Q Δ T and replace Δ Δ with d :

Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.

Example 1.5

Calculating the required heat.

  • Calculate the temperature difference: Δ T = T f − T i = 60.0 ° C . Δ T = T f − T i = 60.0 ° C .
  • Calculate the mass of water. Because the density of water is 1000 kg/m 3 1000 kg/m 3 , 1 L of water has a mass of 1 kg, and the mass of 0.250 L of water is m w = 0.250 kg m w = 0.250 kg .
  • Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3 : Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ . Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ .
  • Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3 : Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ . Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ .
  • Find the total transferred heat: Q Total = Q W + Q Al = 89.8 kJ . Q Total = Q W + Q Al = 89.8 kJ .

Significance

Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of energy had been added with a blowtorch instead of mechanically.)

Example 1.6

Calculating the temperature increase from the work done on a substance.

Calculate the temperature increase of 10 kg of brake material with an average specific heat of 800 J/kg · °C 800 J/kg · °C if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.

Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energy is dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so take Q = M g h / 10 Q = M g h / 10 . Then we calculate the temperature change from the heat transferred, using

where m is the mass of the brake material. Insert the given values to find

In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated from everything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out is called a calorimeter , and the use of a calorimeter to make measurements (typically of heat or specific heat capacity) is called calorimetry .

We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally isolated from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due to conservation of energy:

We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered positive; the heat lost, negative.

Example 1.7

Calculating the final temperature in calorimetry.

  • Use the equation for heat transfer Q = m c Δ T Q = m c Δ T to express the heat transferred from the pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature: Q hot = m A1 c A1 ( T f − 150 ° C ) . Q hot = m A1 c A1 ( T f − 150 ° C ) .
  • Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water, and the final temperature: Q cold = m w c w ( T f − 20.0 ° C ) . Q cold = m w c w ( T f − 20.0 ° C ) .
  • Note that Q hot < 0 Q hot < 0 and Q cold > 0 Q cold > 0 and that as stated above, they must sum to zero: Q cold + Q hot = 0 Q cold = − Q hot m w c w ( T f − 20.0 ° C ) = − m A1 c A1 ( T f − 150 ° C ) . Q cold + Q hot = 0 Q cold = − Q hot m w c w ( T f − 20.0 ° C ) = − m A1 c A1 ( T f − 150 ° C ) .
  • Bring all terms involving T f T f on the left hand side and all other terms on the right hand side. Solving for T f , T f , T f = m A1 c A1 ( 150 ° C ) + m w c w ( 20.0 ° C ) m A1 c A1 + m w c w , T f = m A1 c A1 ( 150 ° C ) + m w c w ( 20.0 ° C ) m A1 c A1 + m w c w , and insert the numerical values: T f = ( 0.500 kg ) ( 900 J/kg ° C ) ( 150 ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) ( 20.0 ° C ) ( 0.500 kg ) ( 900 J/kg ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) = 59.1 ° C . T f = ( 0.500 kg ) ( 900 J/kg ° C ) ( 150 ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) ( 20.0 ° C ) ( 0.500 kg ) ( 900 J/kg ° C ) + ( 0.250 kg ) ( 4186 J/kg ° C ) = 59.1 ° C .

Check Your Understanding 1.3

If 25 kJ is necessary to raise the temperature of a rock from 25 °C to 30 ° C, 25 °C to 30 ° C, how much heat is necessary to heat the rock from 45 °C to 50 ° C 45 °C to 50 ° C ?

Example 1.8

Temperature-dependent heat capacity.

We solve this equation for Q by integrating both sides: Q = m ∫ T 1 T 2 c d T . Q = m ∫ T 1 T 2 c d T .

Then we substitute the given values in and evaluate the integral:

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  • Authors: Samuel J. Ling, William Moebs, Jeff Sanny
  • Publisher/website: OpenStax
  • Book title: University Physics Volume 2
  • Publication date: Oct 6, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/university-physics-volume-2/pages/1-introduction
  • Section URL: https://openstax.org/books/university-physics-volume-2/pages/1-4-heat-transfer-specific-heat-and-calorimetry

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solving calorimetry problems

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Coffee Cup Calorimetry

So how can such simple equipment be used to measure the quantity of heat gained or lost by a system? We have learned on the previous page , that water will change its temperature when it gains or loses energy. And in fact, the quantity of energy gained or lost is given by the equation

Q = m water •C water •ΔT water

where C water is 4.18 J/g/°C. So if the mass of water and the temperature change of the water in the coffee cup calorimeter can be measured, the quantity of energy gained or lost by the water can be calculated.

Q ice = - Q surroundings = -Q calorimeter

The role of the Styrofoam in a coffee cup calorimeter is that it reduces the amount of heat exchange between the water in the coffee cup and the surrounding air. The value of a lid on the coffee cup is that it also reduces the amount of heat exchange between the water and the surrounding air. The more that these other heat exchanges are reduced, the more true that the above mathematical equation will be. Any error analysis of a calorimetry experiment must take into consideration the flow of heat from system to calorimeter to other parts of the surroundings . And any design of a calorimeter experiment must give attention to reducing the exchanges of heat between the calorimeter contents and the surroundings .

Bomb Calorimetry

The coffee cup calorimeters used in high school science labs provides students with a worthwhile exercise in calorimetry. But at the professional level, a cheap Styrofoam cup and a thermometer isn't going to assist a commercial food manufacturer in determining the Calorie content of their products. For situations in which exactness and accuracy is at stake, a more expensive calorimeter is needed. Chemists often use a device known as a bomb calorimeter to measure the heat exchanges associated with chemical reactions, especially combustion reactions. Having little to nothing to do with bombs of the military variety, a bomb calorimeter includes a reaction chamber where the reaction (usually a combustion reaction) takes place. The reaction chamber is a strong vessel that can withstand the intense pressure of heated gases with exploding. The chamber is typically filled with mostly oxygen gas and the fuel . An electrical circuit is wired into the chamber in order to electrically ignite the contents in order to perform a study of the heat released upon combustion. The reaction chamber is surrounded by a jacket of water with a thermometer inserted. The heat released from the chamber warms the water-filled jacket, allowing a scientist to determine the quantity of energy released by the reaction.

solving calorimetry problems

Solving Calorimetry Problems

Now let's look at a few examples of how a coffee cup calorimeter can be used as a tool to answer some typical lab questions. The next three examples are all based on laboratory experiments involving calorimetry.

Example Problem 1: A physics class has been assigned the task of determining an experimental value for the heat of fusion of ice. Anna Litical and Noah Formula dry and mass out 25.8-gram of ice and place it into a coffee cup with 100.0 g of water at 35.4°C. They place a lid on the coffee cup and insert a thermometer. After several minutes, the ice has completely melted and the water temperature has lowered to 18.1°C. What is their experimental value for the specific heat of fusion of ice?

The basis for the solution to this problem is the recognition that the quantity of energy lost by the water when cooling is equal to the quantity of energy required to melt the ice. In equation form, this could be stated as

Q ice = -Q calorimeter

(The negative sign indicates that the ice is gaining energy and the water in the calorimeter is losing energy.) Here the calorimeter (as in the Q calorimeter term) is considered to be the water in the coffee cup. Since the mass of this water and its temperature change are known, the value of Q calorimeter can be determined.

Q calorimeter = m•C•ΔT Q calorimeter = (100.0 g)•(4.18 J/g/°C)•(18.1°C - 35.4°C) Q calorimeter = -7231.4 J

Q ice = m ice •ΔH fusion-ice +7231.4 J = (25.8 g)•ΔH fusion-ice ΔH fusion-ice = (+7231.4 J)/(25.8 g) ΔH fusion-ice = 280.28 J/g ΔH fusion-ice = 2.80x10 2 J/g (rounded to two significant figures)

Example Problem 2: A chemistry student dissolves 4.51 grams of sodium hydroxide in 100.0 mL of water at 19.5°C (in a calorimeter cup). As the sodium hydroxide dissolves, the temperature of the surrounding water increases to 31.7°C. Determine the heat of solution of the sodium hydroxide in J/g.

Once more, the solution to this problem is based on the recognition that the quantity of energy released when sodium hydroxide dissolves is equal to the quantity of energy absorbed by the water in the calorimeter. In equation form, this could be stated as

(The negative sign indicates that the NaOH is losing energy and the water in the calorimeter is gaining energy.) Since the mass and temperature change of the water have been measured, the energy gained by the water (calorimeter) can be determined.

Q calorimeter = m•C•ΔT Q calorimeter = (100.0 g)•(4.18 J/g/°C)•(31.7°C - 19.5°C) Q calorimeter = 5099.6 J

The assumption is that this energy gained by the water is equal to the quantity of energy released by the sodium hydroxide when dissolving. So Q NaOH-dissolving = -5099.6 J. (The negative sign indicates an energy lost.) This quantity is the amount of heat released when dissolving 4.51 grams of the sodium hydroxide. When the heat of solution is determined on a per gram basis, this 5099.6 J of energy must be divided by the mass of sodium hydroxide that is being dissolved.

ΔH solution = Q NaOH-dissolving / m NaOH ΔH solution = (-5099.6 J) / (4.51 g) ΔH solution = -1130.7 J/g ΔH solution = -1.13x10 3 J/g (rounded to three significant figures)

Example Problem 3: A large paraffin candle has a mass of 96.83 gram. A metal cup with 100.0 mL of water at 16.2°C absorbs the heat from the burning candle and increases its temperature to 35.7°C. Once the burning is ceased, the temperature of the water was 35.7°C and the paraffin had a mass of 96.14 gram. Determine the heat of combustion of paraffin in kJ/gram. GIVEN: density of water = 1.0 g/mL.

As is always the case, calorimetry is based on the assumption that all the heat lost by the system is gained by the surroundings . It is assumed that the surroundings is the water that undergoes the temperature change. In equation form, it could be stated that

Q paraffin = -Q water

Since the mass and temperature change of the water are known, the energy gained by the water in the calorimeter can be determined.

Q calorimeter = m•C•ΔT Q calorimeter = (100.0 g)•(4.18 J/g/°C)•(35.7°C - 16.2°C) Q calorimeter = 8151 J

The paraffin released 8151 J or 8.151 kJ of energy when burned. This is based on the burning of 0.69 gram (96.83 g - 96.14 g). To determine the heat of combustion on a per gram basis, the Q paraffin value (-8.151 kJ) must be divided by the mass of paraffin burned:

ΔH combustion - paraffin = (-8.151 kJ) / (0.69 g) ΔH combustion - paraffin = -11.813 kJ/g ΔH combustion - paraffin = -12 kJ/g (rounded to two significant digits)

Check Your Understanding

1. Consider Example Problem 3 above. Identify as many sources of error as you can. For each source indicate the direction of error that would have resulted. That is, identify whether the error would have caused the experimentally derived value to be less than or more than the accepted value.

Answers will vary. Three common choices include: A. Energy is transferred from the water to the surrounding air. This would cause the experimental value to be less than the accepted value since this energy is not contributing to the water's temperature change. B. Energy is being absorbed by the metal cup as the metal also encounters a temperature change. This would cause since this energy is not being accounted for in the calculations. C. Some of the energy released by the burning candle fails to warm either the cup or the water. This energy simply warms the surrounding air. Failure to account for this energy would cause the experimental value to be less than the accepted value.

2. A 2.15-gram cashew nut is burned. The heat released raises the temperature of a 100.0-gram sample of water from 18.2°C to 31.5°C. The mass of the nut after the experiment is 1.78 grams. Determine the calorie content of the nut in Calories/gram. Assume that the water is only able to absorb 25% of the heat released by the burning nut. Given 1.00 Calorie=4.18 kJ.

Answer: ~15 Cal/g

Q water = m water •C water •ΔT water Q water = (100.0 g)•(4.18J/g/°C)•(31.5°C - 18.2°C) = 5559.4 J = 5.5594 kJ Q water = 5.5594 kJ•(1.00 Calorie/4.18 kJ) = 1.3560 Calorie The energy absorbed by the water is one-fourth (25%) of the energy released by the nut. Q nut = -1.3560 Calorie/0.25 = -5.4238 Calorie This 5.4238 Calorie of energy was released by burning 0.37 grams of the Cashew. To determine the Calorie content on a per gram basis, the Calorie-to-gram ratio must be determined. Calorie Content = 5.4248 Calorie/0.37 gram = 14.6589 Cal/g Calorie content = 15 Calorie/gram (rounded to two significant figures)

Calorimetry and Heat Flow: Worked Chemistry Problems

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Calorimetry is the study of heat transfer and changes of state resulting from chemical reactions, phase transitions, or physical changes. The tool used to measure heat change is the calorimeter. Two popular types of calorimeters are the coffee cup calorimeter and bomb calorimeter.

These problems demonstrate how to calculate heat transfer and enthalpy change using calorimeter data. While working these problems, review the sections on coffee cup and bomb calorimetry and the laws of thermochemistry .

Coffee Cup Calorimetry Problem

The following acid-base reaction is performed in a coffee cup calorimeter:

  • H + (aq) + OH - (aq) → H 2 O(l)

The temperature of 110 g of water rises from 25.0 C to 26.2 C when 0.10 mol of H + is reacted with 0.10 mol of OH - .

  • Calculate q water
  • Calculate ΔH for the reaction
  • Calculate ΔH if 1.00 mol OH - reacts with 1.00 mol H +

Use this equation:

  • q = ( specific heat ) x m x Δt

Where q is heat flow, m is mass in grams , and Δt is the temperature change. Plugging in the values given in the problem, you get:

  • q water = 4.18 (J / g·C;) x 110 g x (26.6 C - 25.0 C)
  • q water = 550 J
  • ΔH = -(q water ) = - 550 J

You know that when 0.010 mol of H + or OH - reacts, ΔH is - 550 J:

  • 0.010 mol H + ~ -550 J

Therefore, for 1.00 mol of H + (or OH - ):

  • ΔH = 1.00 mol H + x (-550 J / 0.010 mol H + )
  • ΔH = -5.5 x 10 4 J
  • ΔH = -55 kJ
  • 550 J (Be sure to have two significant figures .)

Bomb Calorimetry Problem

When a 1.000 g sample of the rocket fuel hydrazine, N 2 H 4 , is burned in a bomb calorimeter, which contains 1,200 g of water, the temperature rises from 24.62 C to 28.16 C. If the C for the bomb is 840 J/C, calculate:

  • q reaction  for combustion of a 1-gram sample
  • q reaction  for combustion of one mole of hydrazine in the bomb calorimeter

For a bomb calorimeter, use this equation:

  • q reaction  = -(qwater + qbomb)
  • q reaction  = -(4.18 J / g·C x mwater x Δt + C x Δt)
  • q reaction  = -(4.18 J / g·C x mwater + C)Δt

Where q is heat flow, m is mass in grams, and Δt is the temperature change. Plugging in the values given in the problem:

  • q reaction  = -(4.18 J / g·C x 1200 g + 840 J/C)(3.54 C)
  • q reaction  = -20,700 J or -20.7 kJ

You now know that 20.7 kJ of heat is evolved for every gram of hydrazine that is burned. Using the  periodic table  to get  atomic weights , calculate that one mole of hydrazine, N 2 H 4 , weight 32.0 g. Therefore, for the combustion of one mole of hydrazine:

  • q reaction  = 32.0 x -20.7 kJ/g
  • q reaction  = -662 kJ
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6.2: Calorimetry

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  Learning Objectives

  • Explain the technique of calorimetry
  • Calculate and interpret heat and related properties using typical calorimetry data

One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry . Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.

A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure \(\PageIndex{1}\)). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.

By convention, q is given a negative (-) sign when the system releases heat to the surroundings (exothermic); q is given a positive (+) sign when the system absorbs heat from the surroundings (endothermic).

Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment. This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure \(\PageIndex{2}\)). These easy-to-use “coffee cup” calorimeters allow more heat exchange with their surroundings, and therefore produce less accurate energy values.

Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure \(\PageIndex{3}\)).

Before we practice calorimetry problems involving chemical reactions, consider a simple example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure \(\PageIndex{4}\)). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:

\[q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{5.3.1} \]

This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:

\[q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{5.3.2} \]

The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that \(q_{substance\, M}\) and \(q_{substance\, W}\) are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, \(q_{substance\, M}\) is a negative value and \(q_{substance\, W}\) is positive, since heat is transferred from M to W.

Example \(\PageIndex{1}\): Heat Transfer between Substances at Different Temperatures

A hot 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water is measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).

The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = − heat taken in by water, or:

\[q_\ce{rebar}=−q_\ce{water} \nonumber \]

Since we know how heat is related to other measurable quantities, we have:

\[(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \nonumber \]

Letting f = final and i = initial, in expanded form, this becomes:

\[ c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber \]

The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:

\[ \mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=-(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \nonumber \]

\[\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \nonumber \]

Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.

Exercise \(\PageIndex{1A}\)

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.

The initial temperature of the copper was 335.6 °C.

Exercise \(\PageIndex{1B}\)

A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.

The final temperature (reached by both copper and water) is 38.7 °C.

 This method can also be used to determine other quantities, such as the specific heat of an unknown metal.

Example \(\PageIndex{2}\): Identifying a Metal by Measuring Specific Heat

A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.

Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:

\[q_\ce{metal}=−q_\ce{water} \nonumber \]

In expanded form, this is:

\[c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber \]

Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:

\[\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \nonumber \]

Solving this:

\[\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C} \nonumber \]

Comparing this with values in Table T4, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.

Exercise \(\PageIndex{2}\)

A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).

\(c_{metal}= 0.13 \;J/g\; °C\)

This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.

When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), q reaction , plus the heat absorbed or lost by the solution (the “surroundings”), \(q_{solution}\), must add up to zero:

\[q_\ce{reaction}+q_\ce{solution}=0\ \label{ 5.3.10} \]

This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:

\[q_\ce{reaction}=−q_\ce{solution} \label{5.3.11} \]

This concept lies at the heart of all calorimetry problems and calculations.

Example \(\PageIndex{3}\): Heat Produced by an Exothermic Reaction

When 50.0 mL of 0.10 M HCl( aq ) and 50.0 mL of 1.00 M NaOH( aq ), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?

\[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l) \nonumber \]

To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.

The heat given off by the reaction is equal to that taken in by the solution. Therefore:

\[q_\ce{reaction}=−q_\ce{solution} \nonumber \]

(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and its surroundings.)

Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:

\[q_\ce{solution}=(c×m×ΔT)_\ce{solution} \nonumber \]

To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 × 10 2 g (two significant figures). The specific heat of water is approximately 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:

\[\mathrm{\mathit q_{solution}=(4.184\:J/g\: °C)(1.0×10^2\:g)(28.9°C−22.0°C)=2.89×10^3\:J} \nonumber \]

Finally, since we are trying to find the heat of the reaction, we have:

\[q_\ce{reaction}=−q_\ce{solution}=−2.89×10^3\:J \nonumber \]

The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat.

Exercise \(\PageIndex{3}\)

When 100 mL of 0.200 M NaCl( aq ) and 100 mL of 0.200 M AgNO 3 ( aq ), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?

\(1.34 \times 10^3\; J\); assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water

Thermochemistry of Hand Warmers

When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure \(\PageIndex{5}\)). A common reusable hand warmer contains a supersaturated solution of NaC 2 H 3 O 2 (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable NaC 2 H 3 O 2 quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).

The process \(\ce{NaC2H3O2}(aq)⟶\ce{NaC2H3O2}(s)\) is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC 2 H 3 O 2 redissolves and can be reused.

Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is

\[\ce{2Fe(s) + 3/2 O2(g) ⟶ Fe2O3(s)}.\ n\nonumber \]

Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.

Example \(\PageIndex{4}\): Heat Flow in an Instant Ice Pack

When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (Figure \(\PageIndex{5}\)). When 3.21 g of solid NH 4 NO 3 dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature decreases to 20.3 °C.

Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.

We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:

\[q_\ce{rxn}=−q_\ce{soln} \nonumber \]

with “rxn” and “soln” used as shorthand for “reaction” and “solution,” respectively.

Assuming also that the specific heat of the solution is the same as that for water, we have:

\[\begin{align*} q_\ce{rxn} &=−q_\ce{soln}=−(c×m×ΔT)_\ce{soln}\\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(20.3°C−24.9°C)]}\\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(−4.6°C)]}\\ &+\mathrm{1.0×10^3\:J=+1.0\:kJ} \end{align*}\]

The positive sign for q indicates that the dissolution is an endothermic process.

Exercise \(\PageIndex{4}\)

When a 3.00-g sample of KCl was added to 3.00 × 10 2 g of water in a coffee cup calorimeter, the temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the KCl? What assumptions did you make?

1.33 kJ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water.

If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.

The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter , is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water (Figure \(\PageIndex{6}\)). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is trapped in the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q , such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.

Video \(\PageIndex{1}\): Video of view how a bomb calorimeter is prepared for action.

Example \(\PageIndex{5}\): Bomb Calorimetry

When 3.12 g of glucose, C 6 H 12 O 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample?

The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)

The heat produced by the reaction is absorbed by the water and the bomb:

\[\begin{align*} &q_\ce{rxn}=−(q_\ce{water}+q_\ce{bomb})\\ &=\mathrm{−[(4.184\:J/g\: °C)×(775\:g)×(35.6°C−23.8°C)+893\:J/°C×(35.6°C−23.8°C)]}\\ &=\mathrm{−(38,300\:J+10,500\:J)}\\ &=\mathrm{−48,800\: J=−48.8\: kJ} \end{align*} \nonumber \]

This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.

Exercise \(\PageIndex{5}\)

When 0.963 g of benzene, C 6 H 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases by 8.39 °C. The bomb has a heat capacity of 784 J/°C and is submerged in 925 mL of water. How much heat was produced by the combustion of the glucose sample?

Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. 1 These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories (1 kcal), the amount of energy needed to heat 1 kg of water by 1 °C.

Measuring Nutritional Calories

In your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) = 1000 cal, or 1 kcal. (This is approximately the amount of energy needed to heat 1 kg of water by 1 °C.)

The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure \(\PageIndex{7}\)).

For the example shown in (b), the total energy per 228-g portion is calculated by:

\[\mathrm{(5\:g\: protein×4\:Calories/g)+(31\:g\: carb×4\:Calories/g)+(12\:g\: fat×9\:Calories/g)=252\:Calories} \label{5.3.X} \]

So, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freeze-dried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food.

Today, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy.

Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between the system being studied and its surroundings. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.

  • 1 Francis D. Reardon et al. “The Snellen human calorimeter revisited, re-engineered and upgraded: Design and performance characteristics.” Medical and Biological Engineering and Computing 8 (2006)721–28, The Snellen human calorimeter revisited, re-engineered and upgraded: design and performance characteristics [link.springer.com] .

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Chapter 1 Temperature and Heat

1.4 Heat Transfer, Specific Heat, and Calorimetry

OpenStax and Paula Herrera-Siklody

Learning Objectives

By the end of this section, you will be able to:

  • Explain phenomena involving heat as a form of energy transfer
  • Solve problems involving heat transfer

We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of energy transfer that is caused by a temperature difference, and it can change the temperature of an object. As we learned earlier in this chapter, heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. Heat transfer is fundamental to such everyday activities as home heating and cooking, as well as many industrial processes. It also forms a basis for the topics in the remainder of this chapter.

We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We discuss another way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the study of the relationship of heat and work, which is the basis of engines and refrigerators and the central topic (and origin of the name) of thermodynamics.

Internal Energy and Heat

A thermal system has internal energy (also called thermal energy), which is the sum of the mechanical energies of its molecules. A system’s internal energy is proportional to its temperature. As we saw earlier in this chapter, if two objects at different temperatures are brought into contact with each other, energy is transferred from the hotter to the colder object until the bodies reach thermal equilibrium (that is, they are at the same temperature). No work is done by either object because no force acts through a distance (as we discussed in Work and Kinetic Energy). These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure 1.9 shows an example of heat transfer.

Figure a shows a soda can at temperature T1 and an ice cube, some distance away at temperature T2. T1 is greater than T2. Figure b shows the can and cube in contact with each other. Both are at temperature T prime.

The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say “the heat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy flow, whereas temperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature.

Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is the calorie (cal), defined as the energy needed to change the temperature of 1.00 g of water by [latex]1.00^\circ\text{C}[/latex]—specifically, between [latex]14.5^\circ\text{C}[/latex] and [latex]15.5^\circ\text{C}[/latex], since there is a slight temperature dependence. Also commonly used is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by [latex]1.00^\circ\text{C}[/latex]. Since mass is most often specified in kilograms, the kilocalorie is convenient. Confusingly, food calories (sometimes called “big calories,” abbreviated Cal) are actually kilocalories, a fact not easily determined from package labeling.

Mechanical Equivalent of Heat

It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of a system. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat — the work needed to produce the same effects as heat transfer . In the units used for these two quantities, the value for this equivalence is

1.000 kcal = 4186 J . 1.000 kcal = 4186 J .

We consider this equation to represent the conversion between two units of energy. (Other numbers that you may see refer to calories defined for temperature ranges other than [latex]14.5^\circ\text{C}[/latex] to [latex]15.5^\circ\text{C}[/latex].)

Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can produce the same effects and measuring the mechanical equivalent of heat. It helped establish the principle of conservation of energy. Gravitational potential energy ( U ) was converted into kinetic energy ( K ), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. Joule’s contributions to thermodynamics were so significant that the SI unit of energy was named after him.

solving calorimetry problems

Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although a system has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A well-defined quantity that depends only on the current state of the system, rather than on the history of that system, is known as a state variable. Temperature and internal energy are state variables. To sum up this paragraph, heat and work are not state variables .

Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in the next section, the temperature does not change when a substance changes from one phase to another. An example is the melting of ice, which can be accomplished by adding heat or by doing frictional work, as when an ice cube is rubbed against a rough surface.

Temperature Change and Heat Capacity

We have noted that heat transfer often causes temperature change. Experiments show that with no phase change and no work done on or by the system, the transferred heat is typically directly proportional to the change in temperature and to the mass of the system, to a good approximation. (Below we show how to handle situations where the approximation is not valid.) The constant of proportionality depends on the substance and its phase, which may be gas, liquid, or solid. We omit discussion of the fourth phase, plasma, because although it is the most common phase in the universe, it is rare and short-lived on Earth.

We can understand the experimental facts by noting that the transferred heat is the change in the internal energy, which is the total energy of the molecules. Under typical conditions, the total kinetic energy of the molecules [latex]K_\text{total}[/latex] is a constant fraction of the internal energy (for reasons and with exceptions that we’ll see in the next chapter). The average kinetic energy of a molecule [latex]K_\text{ave}[/latex] is proportional to the absolute temperature. Therefore, the change in internal energy of a system is typically proportional to the change in temperature and to the number of molecules, N . Mathematically, [latex]{\Delta}U \propto \Delta K_\text{total} = N K_\text{ave} \propto N{\Delta}T[/latex] The dependence on the substance results in large part from the different masses of atoms and molecules. We are considering its heat capacity in terms of its mass, but as we will see in the next chapter, in some cases, heat capacities per molecule are similar for different substances. The dependence on substance and phase also results from differences in the potential energy associated with interactions between atoms and molecules.

Heat Transfer and Temperature Change

A practical approximation for the relationship between heat transfer and temperature change is:

Q = m c Δ T , Q = m c Δ T , where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and [latex]{\Delta}T[/latex] is the change in temperature. The symbol c stands for the specific heat (also called “specific heat capacity” ) and depends on the material and phase. The specific heat is numerically equal to the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00°C. The SI unit for specific heat is [latex]J/(kg \times K)[/latex] or [latex]J/(kg \times ^{\circ}C)[/latex]. (Recall that the temperature change [latex]{\Delta}T[/latex] is the same in units of kelvin and degrees Celsius.)

Values of specific heat must generally be measured, because there is no simple way to calculate them precisely. Table 1.3 lists representative values of specific heat for various substances. We see from this table that the specific heat of water is five times that of glass and 10 times that of iron, which means that it takes five times as much heat to raise the temperature of water a given amount as for glass, and 10 times as much as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.

The specific heats of gases depend on what is maintained constant during the heating—typically either the volume or the pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the second (in parentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic theory of gases.

In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given in terms of an infinitesimal change in temperature. To do this, we note that [latex]c = \frac{1}{m}\frac{{\Delta}Q}{{\Delta}T}[/latex] and replace [latex]\Delta[/latex] with d :

c = 1 m d Q d T . c = 1 m d Q d T .

Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.

Example 1.5

Calculating the Required Heat A 0.500-kg aluminum pan on a stove and 0.250 L of water in it are heated from [latex]20.0^\circ\text{C}[/latex] to [latex]80.0^\circ\text{C}[/latex]. (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?

Strategy We can assume that the pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and that of the pan are increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 1.3.

  • Calculate the temperature difference: Δ T = T f − T i = 60.0 ° C . Δ T = T f − T i = 60.0 ° C .
  • Calculate the mass of water. Because the density of water is [latex]{1000}\text{ kg} /m^{3}[/latex], 1 L of water has a mass of 1 kg, and the mass of 0.250 L of water is [latex]m_w = {0.250} \text{ kg}[/latex].
  • Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3: Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ . Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ .
  • Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3: Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ . Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ .
  • Find the total transferred heat: Q Total = Q W + Q Al = 89.8 kJ . Q Total = Q W + Q Al = 89.8 kJ .

Significance In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times that of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water as for the aluminum pan.

Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of energy had been added with a blowtorch instead of mechanically.)

Example 1.6

Calculating the Temperature Increase from the Work Done on a Substance Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material (Figure 1.11). This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. Since the mass of the truck is much greater than that of the brake material absorbing the energy, the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat.

solving calorimetry problems

Calculate the temperature increase of 10 kg of brake material with an average specific heat of [latex]800J/ \text{ kg} \cdot ^\circ {C}[/latex] if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.

Strategy We calculate the gravitational potential energy (Mgh) that the entire truck loses in its descent, equate it to the increase in the brakes’ internal energy, and then find the temperature increase produced in the brake material alone.

Solution First we calculate the change in gravitational potential energy as the truck goes downhill:

M g h = ( 10,000 kg ) ( 9.80 m/s 2 ) ( 75.0 m ) = 7.35 × 10 6 J . M g h = ( 10,000 kg ) ( 9.80 m/s 2 ) ( 75.0 m ) = 7.35 × 10 6 J . Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energy is dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so take [latex]Q = Mgh/10[/latex]. Then we calculate the temperature change from the heat transferred, using

Δ T = Q m c , Δ T = Q m c , where m is the mass of the brake material. Insert the given values to find

Δ T = 7.35 × 10 5 J ( 10 kg ) ( 800 J/kg °C ) = 92 ° C . Δ T = 7.35 × 10 5 J ( 10 kg ) ( 800 J/kg °C ) = 92 ° C . Significance If the truck had been traveling for some time, then just before the descent, the brake temperature would probably be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material very high, so this technique is not practical. Instead, the truck would use the technique of engine braking. A different idea underlies the recent technology of hybrid and electric cars, where mechanical energy (kinetic and gravitational potential energy) is converted by the brakes into electrical energy in the battery, a process called regenerative braking.

In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated from everything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out is called a calorimeter , and the use of a calorimeter to make measurements (typically of heat or specific heat capacity) is called calorimetry .

We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally isolated from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due to conservation of energy:

We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered positive; the heat lost, negative.

Example 1.7

Calculating the Final Temperature in Calorimetry Suppose you pour 0.250 kg of [latex]20.0-^\circ\text{C}[/latex] water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of [latex]150^\circ\text{C}[/latex]. Assume no heat transfer takes place to anything else: The pan is placed on an insulated pad, and heat transfer to the air is neglected in the short time needed to reach equilibrium. Thus, this is a calorimetry problem, even though no isolating container is specified. Also assume that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium?

Strategy Originally, the pan and water are not in thermal equilibrium: The pan is at a higher temperature than the water. Heat transfer restores thermal equilibrium once the water and pan are in contact; it stops once thermal equilibrium between the pan and the water is achieved. The heat lost by the pan is equal to the heat gained by the water—that is the basic principle of calorimetry.

  • Use the equation for heat transfer [latex]Q = mc{\Delta}T[/latex] to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature: [latex]Q_\text{hot} = m_\text{A1} c_\text{A1} ( T_\text{f} - 150^\circ\text{C}) .[/latex]
  • Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water, and the final temperature: [latex]Q_\text{cold} = m_\text{w} c_\text{w} ( T_\text{f} - 20.0^\circ \text{C}) .[/latex]
  • Note that [latex]Q_\text{hot} 0[/latex] and [latex]Q_\text{cold} > 0[/latex] and that as stated above, they must sum to zero: Q cold + Q hot = 0 Q cold = − Q hot m w c w ( T f − 20.0 ° C ) = − m A1 c A1 ( T f − 150 ° C ) . Q cold + Q hot = 0 Q cold = − Q hot m w c w ( T f − 20.0 ° C ) = − m A1 c A1 ( T f − 150 ° C ) .
  • This a linear equation for the unknown final temperature, [latex]T_f[/latex]. Solving for [latex]T_f[/latex], [latex]T_\text{f} = \frac{m_{A1}c_{A1}(150^\circ \text{C}) + {m_\text{w} c_\text{w}}(20.0^\circ \text{C})}{{m_\text{A1} c_\text{A1}}+{m_\text{w} c_\text{w}}}[/latex]

and insert the numerical values:

[latex]T_\text{f} = \frac{(0.500 \text{kg}) (900 \text{J/kg} ^\circ \text{C}) (150 ^\circ \text{C}) + (0.250 \text{kg}) (4186 \text{J/kg} ^\circ \text{C})(20.0 ^\circ \text{C})}{(0.500 \text{kg}) (900 \text{J/kg} ^\circ \text{C}) + (0.250 \text{kg})(4186 \text{J/kg}^\circ\text{C})} = 59.1^\circ \text{C}[/latex]

Significance Why is the final temperature so much closer to [latex]20.0^\circ\text{C}[/latex] than to [latex]150^\circ\text{C}[/latex]? The reason is that water has a greater specific heat than most common substances and thus undergoes a smaller temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during the day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter).

Check Your Understanding 1.3

If 25 kJ is necessary to raise the temperature of a rock from [latex]25^\circ\text{C}[/latex] to [latex]30^\circ\text{C}[/latex], how much heat is necessary to heat the rock from [latex]45^\circ\text{C}[/latex] to [latex]50^\circ\text{C}[/latex]?

Example 1.8

Temperature-Dependent Heat Capacity At low temperatures, the specific heats of solids are typically proportional to [latex]T^3[/latex]. The first understanding of this behavior was due to the Dutch physicist Peter Debye, who in 1912, treated atomic oscillations with the quantum theory that Max Planck had recently used for radiation. For instance, a good approximation for the specific heat of salt, NaCl, is [latex]c = 3.33 \times {10}^4 \frac{\text{J}}{\text{kg} \cdot \text{k}} {\bigg(\frac{T}{321\text{K}}\bigg)}^3[/latex]. The constant 321 K is called the Debye temperature of NaCl, [latex]{\Theta}_\text{D}[/latex], and the formula works well when [latex]T 0.04{\Theta}_\text{D}[/latex]. Using this formula, how much heat is required to raise the temperature of 24.0 g of NaCl from 5 K to 15 K? Solution

Because the heat capacity depends on the temperature, we need to use the equation

[latex]c = \frac{1}{m}\frac{dQ}{dT} .[/latex]

We solve this equation for Q by integrating both sides: [latex]Q = m \int_{T_1}^{T_2} cdT .[/latex]

Then we substitute the given values in and evaluate the integral:

[latex]Q = (0.024kg) \int_{T_1}^{T_2} 333 \times {10}^4 \frac{J}{kg \cdot K} {\bigg(\frac{T}{321K}\bigg)}^3 dT = {\bigg(6.04 \times {10}^{-4} \frac{J}{K^4}\bigg) T^4 |}_{5K}^{15K} = 30.2J .[/latex]

Significance If we had used the equation [latex]Q=mc{\Delta}T[/latex] and the room-temperature specific heat of salt, [latex]880\text{J/kg} \cdot \text{K}[/latex], we would have gotten a very different value.

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1.4 Heat Transfer, Specific Heat, and Calorimetry Copyright © 2016 by OpenStax and Paula Herrera-Siklody is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Calorimetry Problems

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  • Physical , Physical Chemistry
  • November 12, 2023

Calorimetry Problems

Imagine you’re in a chemistry lab, conducting an experiment. As you mix different substances together, you notice a sudden release of heat. Curiosity piqued, you wonder how much heat was generated and what it means for the reaction. This is where calorimetry comes into play.

Calorimetry is the science of measuring heat transfer in chemical reactions, providing valuable insights into energy changes. It helps us understand the thermal energy involved when substances react with one another. Accurate calorimetry measurements are crucial for both research and industrial applications, aiding in determining the enthalpy change of a reaction.

Common Problems Encountered in Calorimetry Experiments

Heat loss or gain to the surroundings.

One common problem encountered in calorimetry experiments is heat loss or gain to the surroundings, which can significantly affect the accuracy of measurements.

When conducting an experiment, it’s essential to ensure that the system is well-insulated to minimize any heat exchange with the environment. This can be achieved by using proper insulation materials and ensuring a tight seal around the calorimeter.

Incomplete Combustion and Accuracy

Another issue that can lead to inaccurate results is incomplete combustion. During a combustion reaction, if there isn’t enough oxygen available for complete combustion, it can result in incomplete combustion products being produced.

These incomplete products may have different energy values than expected, leading to errors in calculating the heat released or absorbed during the reaction. To mitigate this problem, it’s crucial to ensure sufficient oxygen supply and proper mixing of reactants.

Mixing Errors during Reactant Preparation

Mixing errors during reactant preparation can also impact the accuracy of calorimetry measurements. If reactants are not thoroughly mixed before initiating a reaction, there might be localized variations in temperature throughout the mixture. This inconsistency can lead to incorrect calculations of enthalpy changes.

To avoid this problem, it’s important to mix reactants thoroughly and ensure uniformity before starting any calorimetry experiment.

Insufficient Insulation and Heat Exchange

Insufficient insulation causes heat exchange, compromising accuracy. Poorly insulated calorimeters lead to inaccurate readings due to heat escaping or entering. Proper insulation with materials like Styrofoam is crucial to minimize heat transfer.

Addressing common calorimetry problems, such as heat loss or gain, incomplete combustion, mixing errors, and insufficient insulation, enhances measurement accuracy. Avoiding these issues ensures precise enthalpy change calculations, providing scientists with accurate data for meaningful conclusions and informed decisions.

Basics of Solving Calorimetry Problems

To solve calorimetry problems, use the equation q = mcΔT, where mass (m), specific heat capacity (c), and temperature change (ΔT) are crucial variables. Plug in consistent units for accurate calculations, ensuring precise determination of energy absorbed or released by substances.

Let’s break down each variable:

Mass (m): This refers to the quantity of matter present in a substance. It is typically measured in grams or kilograms.

Specific heat capacity (c): This property represents how much heat energy a substance can absorb or release per unit mass. It is measured in J/g°C or J/kg°C.

Temperature change (ΔT): This indicates the difference between initial and final temperatures. It is usually measured in degrees Celsius or Kelvin.

Understanding these variables and their significance within the calorimetry equation equips you better to tackle problems involving heat transfer and energy changes during chemical reactions or physical processes.

Remember, practicing with calorimetry problems and applying this equation increases your comfort and accuracy in solving them.

Step-by-Step Guide for Calculating Calorimetry

To analyze calorimetry problems accurately, determine the system and surroundings for a clear understanding of heat transfer. Identify crucial known values such as mass, specific heat capacity, and initial/final temperatures.

Calculate the temperature change (∆T) by subtracting initial from final temperatures. This value indicates the temperature change during the process.

Apply the q = mc∆T equation to find the heat transferred. Plug in known values (mass, specific heat capacity, and ∆T) to calculate q, representing the heat energy gained or lost by a substance.

Here’s an example to illustrate these steps:

Let’s say we have a calorimeter with a known mass of 50 grams. The specific heat capacity of water is 4.18 J/g°C, and we measure an initial temperature of 25°C and a final temperature of 45°C.

Determine that our system is the calorimeter and its contents.

Identify known values: mass = 50g, specific heat capacity = 4.18 J/g°C, initial temperature = 25°C, final temperature = 45°C.

Calculate ∆T: ∆T = Final Temperature – Initial Temperature = 45°C – 25°C = 20°C.

Apply q = mc∆T: q = (50g)(4.18 J/g°C)(20°C) ≈ 4180 J.

By following these steps, we are able to calculate that approximately 4180 Joules of heat energy were transferred in this particular calorimetry problem.

Advanced Techniques for Complex Calorimetry Problem Solving

Enthalpy of formation calculations.

Calculating the enthalpy of formation can become more complex when dealing with multiple reactants and products, each with different stoichiometric coefficients. This means that the heat released or absorbed during a reaction might not be as straightforward to determine.

To tackle this challenge, one useful technique is to apply Hess’s Law . This law allows us to calculate enthalpy changes by using known enthalpies of other reactions involved in the process. By combining these known values, we can determine the overall enthalpy change for the desired reaction.

Using Hess’s Law

Hess’s Law involves manipulating chemical equations and their corresponding enthalpies to obtain the desired reaction equation. Here’s how it works:

Identify known reactions: Find other reactions with known enthalpies that involve some of the same substances as your target reaction.

Manipulate equations: Multiply or reverse equations as needed so that they align with your target reaction.

Adjust enthalpies: Multiply the enthalpies by appropriate factors based on any changes made to the equations.

Combine equations: Add or subtract the manipulated equations together, canceling out any common substances.

Calculate overall enthalpy change: The sum of the adjusted enthalpies from step 3 gives you the overall change in enthalpy for your target reaction.

Application Examples

Let’s explore some examples where advanced calorimetry techniques come into play:

Combustion Reactions: Determining the heat released during combustion processes can involve multiple reactants and products, making it an ideal scenario for applying Hess’s Law.

Solution Reactions: When dissolving substances in water, calculating heat changes can become intricate due to different stoichiometric coefficients and varying concentrations.

Microwave Calorimetry: Advanced techniques are employed in microwave calorimetry to measure heat changes during chemical reactions that occur under microwave radiation.

Coal Combustion: Analyzing the enthalpy changes in coal combustion involves intricate calculations due to the complex composition of coal and its combustion products.

Unknown Metal Reactions: When determining the identity of an unknown metal through calorimetry, advanced techniques are necessary to account for multiple reactants and products involved in the reaction.

Remember, these advanced techniques may require practice and a solid understanding of stoichiometry. Don’t hesitate to seek additional resources or consult with experts like Varsity Tutors to enhance your problem-solving skills in calorimetry.

Tips and Tricks for Accurate Calorimetry Calculations

Minimize heat loss/gain with proper insulation techniques.

To ensure accurate calorimetry calculations, it is crucial to minimize heat loss or gain during the experiment. One effective method is to use proper insulation techniques, such as a well-designed calorimeter. This will help maintain a stable environment and prevent external factors from affecting the results.

Ensure uniform temperature distribution with stirring

Another important aspect of accurate calorimetry calculations is ensuring a uniform temperature distribution within the system. To achieve this, it is recommended to use a stirring device during the experiment. The stirring action promotes better mixing of substances, allowing for more consistent temperature readings.

Account for the heat capacity of the calorimeter

When calculating heat transfer in calorimetry experiments, it’s essential to account for the heat capacity of the calorimeter itself. The calorimeter absorbs some of the heat energy exchanged between substances, affecting the overall calculation. By considering its heat capacity in your calculations, you can obtain more precise results.

Regularly calibrate equipment for accuracy

Calibration plays a vital role in maintaining accuracy in calorimetry experiments. It involves checking and adjusting instruments and equipment to ensure they provide reliable measurements. Regularly calibrating your equipment helps identify any potential errors or deviations and allows you to make necessary adjustments before conducting experiments.

By following these tips and tricks, you can enhance the accuracy of your calorimetry calculations and obtain more reliable data.

Real-World Applications and Conclusion

Congratulations! You’ve made it through the various sections on calorimetry problems, from understanding the basics to learning advanced techniques. Now that you have a solid foundation, let’s explore some real-world applications of calorimetry and wrap up our journey.

Calorimetry, with practical applications in various fields, determines nutritional content in the food industry. Scientists calculate calories by measuring energy released during combustion, aiding informed dietary choices for consumers.

In conclusion, mastering calorimetry problems opens doors to countless opportunities for understanding energy transfer and making informed decisions in various industries. So go ahead and apply your newfound knowledge to solve real-world challenges. Remember, practice makes perfect! Keep honing your skills by tackling more complex problems and exploring additional resources available online. Happy problem-solving!

Can I use calorimetry to measure heat changes in chemical reactions?

Yes! Calorimetry is commonly used to measure heat changes in chemical reactions. By conducting experiments with a calorimeter, you can determine the amount of heat absorbed or released during a reaction, providing valuable insights into its thermodynamic properties.

Is there any difference between bomb calorimetry and solution calorimetry?

Absolutely! Bomb calorimetry involves measuring heat changes in reactions occurring under constant volume conditions, typically involving combustion processes. On the other hand, solution calorimetry measures heat changes when reactants are dissolved or mixed together under constant pressure conditions.

Are there any safety precautions I should take while performing calorimetric experiments?

Safety should always be a priority when working with any experimental setup involving heat sources or potentially hazardous substances. Ensure proper ventilation, wear appropriate protective gear, and follow established safety protocols. It is also crucial to be aware of the specific risks associated with the chemicals or equipment you are using.

Can calorimetry be used in environmental studies?

Absolutely! In environmental studies, researchers apply calorimetry to measure heat changes linked to processes like combustion or thermal decomposition. This information aids in comprehending the impact of these processes on energy consumption, pollution levels, and climate change.

Are there any software or online tools available for solving calorimetry problems?

Yes, several software programs and online tools are available that can assist in solving calorimetry problems. These tools often provide step-by-step calculations and allow you to input variables easily. However, it’s important to understand the underlying principles and concepts before relying solely on these tools for problem-solving.

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  1. 8.5.1: Practice Problems- Calorimetry

    PROBLEM 8.5.1. 7. The addition of 3.15 g of Ba (OH) 2 •8H 2 O to a solution of 1.52 g of NH 4 SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by ...

  2. 8.2: Calorimetry (Problems)

    Answer. PROBLEM 8.2.6 8.2. 6. When 50.0 g of 0.200 M NaCl ( aq) at 24.1 °C is added to 100.0 g of 0.100 M AgNO 3 ( aq) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl ( s) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced.

  3. Calorimetry Practice Problems

    Calorimetry Practice Problems In this set of practice questions, we will go over the main types of questions on calorimetry including the heat capacity, the heat of reaction, finding the final temperature of a mixture, constant pressure calorimetry, and constant-volume calorimetry.

  4. How To Solve Basic Calorimetry Problems in Chemistry

    2.2K 220K views 6 years ago New AP & General Chemistry Video Playlist This chemistry video tutorial explains how to solve basic calorimetry problems. It discusses how to calculate the heat energy...

  5. Calorimetry Problems, Thermochemistry Practice, Specific Heat Capacity

    This chemistry video tutorial explains how to solve calorimetry problems in thermochemistry. It shows you how to calculate the quantity of heat transferred ...

  6. Heat capacity and calorimetry (practice)

    A The final temperature is closer to T 1 than to T 2 . The final temperature is exactly halfway between T 1 and T 2 . B The final temperature is exactly halfway between T 1 and T 2 . The final temperature is closer to T 2 than to T 1 . C The final temperature is closer to T 2 than to T 1 .

  7. Solving a Basic Calorimetry Problem

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  8. 5.2 Calorimetry

    A thermometer and stirrer extend through the cover into the reaction mixture. Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer.

  9. 7.3: Heats of Reactions and Calorimetry

    Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). ... Solving this gives Ti,rebar= 248 °C, so the ...

  10. 10.2 Calorimetry

    Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). ... Solving this gives T i,rebar = 248 °C, so the ...

  11. How To Solve Basic Calorimetry Problems in Chemistry

    Intro to Projectile Motion: Horizontal Launch 35m. Negative (Downward) Launch 25m. Symmetrical Launch 25m. Projectiles Launched From Moving Vehicles 15m. Special Equations in Symmetrical Launches 16m. Positive (Upward) Launch 42m. Using Equation Substitution 17m. 6. Intro to Forces (Dynamics) 3h 22m.

  12. 1.4 Heat Transfer, Specific Heat, and Calorimetry

    An important idea in solving calorimetry problems is that during a heat transfer between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due to conservation of energy: Q cold + Q hot = 0. Q cold + Q hot = 0. 1.6.

  13. Calorimeters and Calorimetry

    Q = mwater•Cwater•ΔTwater where Cwater is 4.18 J/g/°C. So if the mass of water and the temperature change of the water in the coffee cup calorimeter can be measured, the quantity of energy gained or lost by the water can be calculated.

  14. Calorimetry and Heat Flow: Worked Chemistry Problems

    Solution Use this equation: q = ( specific heat) x m x Δt Where q is heat flow, m is mass in grams, and Δt is the temperature change. Plugging in the values given in the problem, you get: q water = 4.18 (J / g·C;) x 110 g x (26.6 C - 25.0 C) q water = 550 J ΔH = - (q water) = - 550 J

  15. Final Temperature Calorimetry Practice Problems

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  16. 6.2: Calorimetry

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  17. Intro to Calorimetry Video Tutorial & Practice

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  18. PDF Hints on solving calorimetry probs 2

    2. q. rxn. = - q. calorimeter. 3. Divide qrxn by the number of moles of the limiting reactant. Multiply by the coefficient of the LR in the balanced equation to get ∆E for the reaction as written. For either of these types of calorimetry (coffee-cup or bomb), once you know ∆H or ∆E, you can easily calculate the other.

  19. 1.4 Heat Transfer, Specific Heat, and Calorimetry

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  20. Calorimetry Problems

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  22. ALEKS: Solving A Basic Calorimetry Problem

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