Heat Capacities

I.                    Introduction

Heat capacity is a term in physics that describes how much heat must be added to a substance to raise its temperature by 1 degree Celsius. This is related to, but distinct from, specific heat, which is the amount of heat needed to raise exactly 1 gram (or some other fixed unit of mass) of a substance by 1 degree Celsius. Deriving a substance's heat capacity C from its specific heat S is a matter of multiplying by the amount of the substance that is present and making sure you are using the same units of mass throughout the problem. Heat capacity, in plain terms, is an index of an object's ability to resist being warmed by the addition of heat energy.

I.                    Pre-test

 1.    The specific heat capacity of aluminum is 0.897 J/g°C.   Determine the amount of heat released when 10.5g of aluminum cools from 240°C to 25°C.

2.    The specific heat capacity of aluminum is 0.897 J/g°C.   What mass of aluminum can be heated from 33°C to 99°C using 450J of heat?

3.    84.0g of a metal are heated to 112ºC, and then placed in a coffee cup calorimeter containing 60.0g of water at 32ºC.  The final temperature in the calorimeter is 42ºC.  What is the specific heat of the metal?

4.    Determine the amount of heat needed to raise the temperature of 15.0g of lead from 22°C to 68°C.  The specific heat capacity of lead is 0.13 J/g°C.

5.    What mass of water can be heated from 15°C to 75°C using 6500J of heat?  The specific heat of water is 4.18 J/g°C.

6.    What mass of lead can be heated from 23°C to 44°C using 125J of heat?  The specific heat capacity of lead is 0.129 J/g°C.

7.    When 32.0g of a substance cools from 85°C to room temperature (25°C), 2400J of heat are released.  Find the specific heat capacity of the substance.

8.    A 16-g piece of iron absorbs 1090 joules of heat energy, and its temperature changes from 25ºC to 175ºC. Calculate the heat capacity of iron.

9.    How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22ºC to 55ºC, if the specific heat of aluminum is 0.90 J/gºC?

10.    To what temperature will a 50.0 g piece of granite raise if it absorbs 5275 joules of heat and its heat capacity is 0.50 J/gºC? The initial temperature of the granite is 20.0ºC.

 

II.                  Discussion

 

Heat Capacities of Gases

The basis of our analysis is that heat is energy in transit. When we add heat to a substance, we are increasing its molecular energy. In this discussion the volume of the gas will remain constant so that we don’t have to worry about energy transfer through mechanical work. If we were to let the gas expand, it would do work by pushing on moving walls of its container, and this additional energy transfer would have to be included in our calculations.             For now, with the volume held constant, we are concerned with the molar heat capacity at constant volume.

In the simple kinetic-molecular model, the molecular energy consists only of the translational kinetic energy of the pointlike molecules. This energy is directly proportional to the absolute temperature T. When the temperature changes by a small amount dT, the corresponding change in kinetic energy is Ktr= 3/2 nRT. When the temperature changes by a small amount dT, the corresponding change in kinetic energy is

From the definition of molar heat capacity at constant volume, we also have

where dQ is the heat input needed for a temperature change dT. Now if Ktr  represents the total molecular energy, as we have assumed, then dQ and dKtr must be equal.

This surprisingly simple result says that the molar heat capacity at constant volume is 3R/2 for any gas whose molecules can be represented as points.

For comparison, Table 18.1 gives measured values of for several gases. We see that for monatomic gases our prediction is right on the money, but that it is way off for diatomic and polyatomic gases.

This comparison tells us that our point-molecule model is good enough for monatomic gases but that for diatomic and polyatomic molecules we need something more sophisticated. For example, we can picture a diatomic molecule as two point masses, like a little elastic dumbbell, with an interaction force between the atoms of the kind shown in Fig. 18.8. Such a molecule can have additional kinetic energy associated with rotation about axes through its center of mass. The atoms may also have vibrating motion along the line joining t hem, with additional kinetic and potential energies.

When heat flows into a monatomic gas at constant volume, all of the added energy goes into an increase in random translational molecular kinetic energy. But when the temperature is increased by the same amount in a diatomic or polyatomic gas, additional heat is needed to supply the increased rotational and vibrational energies. Thus polyatomic gases have larger molar heat capacities than monatomic gases, as Table 18.1 shows.

But how do we know how much energy is associated with each additional kind of motion of a complex molecule, compared to the translational kinetic energy? The new principle that we need is called the principle of equipartition of energy. It can be derived from sophisticated statistical-mechanics considerations; that derivation is beyond our scope, and we will treat the principle as an axiom.

The principle of equipartition of energy states that each velocity component (either linear or angular) has, on average, an associated kinetic energy per molecule of or one-half the product of the Boltzmann constant and the absolute temperature. The number of velocity components needed to describe the motion of a molecule completely is called the number of degrees of freedom. For a monatomic gas, there are three degrees of freedom (for the velocity components and this gives a total average kinetic energy per molecule of 3(1/2 kT )consistent with Eq. (18.16).

For a diatomic molecule there are two possible axes of rotation, perpendicular to each other and to the molecule’s axis. (We don’t include rotation about the molecule’s own axis because in ordinary collisions there is no way for this rotational motion to change.) If we assign five degrees of freedom to a diatomic molecule, the average total kinetic energy per molecule is instead of The total kinetic energy of n moles is   

and the molar heat capacity (at constant volume) is  

This agrees within a few percent with the measured values for diatomic gases given in Table 18.1. Vibrational motion can also contribute to the heat capacities of gases. Molecular bonds are not rigid; they can stretch and bend, and the resulting vibrations lead to additional degrees of freedom and additional energies. For most diatomic gases, however, vibrational motion does not contribute appreciably to heat capacity. The reason for this is a little subtle and involves some concepts of quantum mechanics. Briefly, vibrational energy can change only in finite steps. If the energy change of the first step is much larger than the energy possessed by most molecules, then nearly all the molecules remain in the minimum-energy state of motion. In that case, changing the temperature does not change their average vibrational energy appreciably, and the vibrational degrees of freedom are said to be “frozen out.” In more complex molecules the gaps between permitted energy levels are sometimes much smaller, and then vibration does contribute to heat capacity. The rotational energy of a molecule also changes by finite steps, but they are usually much smaller; the “freezing out” of rotational degrees of freedom occurs only in rare instances, such as for the hydrogen molecule below about 100 K.

In Table 18.1 the large values of Cv for some polyatomic molecules show the contributions of vibrational energy. In addition, a molecule with three or more atoms that are not in a straight line has three, not two, rotational degrees of freedom.

From this discussion we expect heat capacities to be temperature-dependent, generally increasing with increasing temperature. Figure 18.19 is a graph of the temperature dependence of for hydrogen gas showing the temperatures at which the rotational and vibrational energies begin to contribute.

Heat Capacities of Solids

 We can carry out a similar heat-capacity analysis for a crystalline solid. Consider a crystal consisting of N identical atoms (a monatomic solid). Each atom is bound to an equilibrium position by interatomic forces. The elasticity of solid materials shows us that these forces must permit stretching and bending of the bonds. We can think of a crystal as an array of atoms connected by little springs (Fig. 18.20). Each atom can vibrate about its equilibrium position.

Each atom has three degrees of freedom, corresponding to its three components of velocity. According to the equipartition principle, each atom has an average kinetic energy of for each degree of freedom. In addition, each atom has potential energy associated with the elastic deformation. For a simple harmonic oscillator (discussed in Chapter 14) it is not hard to show that the average kinetic energy of ½ kT an atom is equal to its average potential energy. In our model of a crystal, each atom is essentially a three-dimensional harmonic oscillator; it can be shown that the equality of average kinetic and potential energies also holds here, provided that the “spring” forces obey Hooke’s law.

Thus we expect each atom to have an average kinetic energy 3/2 kT and an average potential energy or an average total energy 3kT per atom. If the crystal contains N atoms or n moles, its total energy is

At low temperatures, the heat capacities of most solids decrease with decreasing temperature (Fig. 18.21) for the same reason that vibrational degrees of freedom of molecules are frozen out at low temperatures. At very low temperatures the quantity kT is much smaller than the smallest energy step the vibrating atoms can take.

Hence most of the atoms remain in their lowest energy states because the next higher energy level is out of reach. The average vibrational energy per atom is then less than 3kT, and the heat capacity per molecule is less than 3k. At higher temperatures when kT is large in comparison to the minimum energy step, the equipartition principle holds, and the total heat capacity is 3k per molecule or 3R per mole as the Dulong and Petit rule predicts.

Quantitative understanding of the temperature variation of heat capacities was one of the triumphs of quantum mechanics during its initial development in the 1920s.

 

I.                    Post-Test

 

1.       It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g·°C?

2.       Calculate the amount of heat released when 7.40 g of water cools from 49° to 28°C.

3.       A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.

4.       A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of the metal?

5.       A hot 1 kg chunk of copper is allowed to cool to 100°C. If the copper gave off 231 kJ of energy, what was the initial temperature of the copper? The specific heat of copper is 0.385 J/g°C.

6.       A 15.0g15.0g piece of cadmium metal absorbs 134J134J of heat while rising from 24.0oC24.0oC to 62.7oC62.7oC. Calculate the specific heat of cadmium.

7.       What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow?

8.       A 10.3 g sample of a reddish-brown metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. What is the specific heat of the metal?

9.       The specific heat capacity of aluminum is 0.897 J/g°C.   Determine the amount of heat released when 10.5g of aluminum cools from 240°C to 25°C.

10.   The specific heat capacity of aluminum is 0.897 J/g°C.   What mass of aluminum can be heated from 33°C to 99°C using 450J of heat?

SOLUTIONS(Post-Test)

1. It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g·°C?
Solution:
Use the formula
q = mcΔT
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature
Putting the numbers into the equation yields:

487.5 J = (25 g)c(75 °C - 25 °C)
487.5 J = (25 g)c(50 °C)
Solve for c:
c = 487.5 J/(25g)(50 °C)
c = 0.39 J/g·°C

Answer:
The specific heat of copper is 0.39 J/g·°C.

2.            Calculate the amount of heat released when 7.40 g of water cools from 49° to 28°C.

q = m × C × ΔT

q = 7.40 g × 4.184 J/g∙°C × (28°C – 49°C)

q = 7.40 g × 4.184 J/g∙°C × (-21°C)

q = -6.5×102 J

 

3.            Question: A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.

Solution: First, let’s the variables we know.

m = 500 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C

Plug these values into the specific heat equation from above.

Q = mcΔT

Q = (500 grams)·(0.129 J/g°C)·(50 °C)

Q = 3225 J

Answer: It took 3225 Joules of energy to heat the lead cube from 25 °C to 75 °C.

 

 

4.  A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of the metal?

Solution: List the information we know.

m = 25 grams
ΔT = 200 °C
Q = 2330 J

Place these into the specific heat equation.

Q = mcΔT

2330 J = (25 g)c(200 °C)

2330 J = (5000 g°C)c

Divide both sides by 5000 g°C

c = 0.466 J/g°C

Answer: The specific heat of the metal is 0.466 J/g°C.

 

 

5.            Question: A hot 1 kg chunk of copper is allowed to cool to 100°C. If the copper gave off 231 kJ of energy, what was the initial temperature of the copper? The specific heat of copper is 0.385 J/g°C.

Solution: List our given variables:

m = 1 kg
Tfinal = 100 °C
Q = -231 kJ (The negative sign is because the copper is cooling and losing energy.)
c = 0.385 J/g°C

We need to make our units consistent with the specific heat units, so let’s convert the mass and energy units.

m = 1 kg = 1000 grams

1 kJ = 1000 J
Q = -231 kJ · (1000 J/kJ) = -231000 J

Plug these values into the specific heat formula.

Q = mcΔT

-231000 J = 1000 g · (0.385 J/g°C) · ΔT

-231000 J = 385 J/°C · ΔT

ΔT = -600 °C

ΔT = (Tfinal – Tinitial)

Plug in the values for ΔT and Tfinal.

-600 °C = (100 °C – Tinitial)

Subtract 100 °C from both sides of the equation.

-600 °C – 100 °C =  – Tinitial

-700 °C = – Tinitial

Tinitial = 700 °C

Answer: The initial temperature of the copper chunk was 700 °C.

 

 

 

6.            A 15.0g15.0g piece of cadmium metal absorbs 134J134J of heat while rising from 24.0oC24.0oC to 62.7oC62.7oC. Calculate the specific heat of cadmium.

Solution:

Step 1: List the known quantities and plan the problem.

Known

• Heat =q=134J=q=134J

• Mass =m=15.0g=m=15.0g

• ΔT=62.7oC−24.0oC=38.7oCΔT=62.7oC−24.0oC=38.7oC

Unknown

• cpcp of cadmium =?J/goC=?J/goC

The specific heat equation can be rearranged to solve for the specific heat.

Step 2: Solve.

cp=qm×ΔT=134J15.0g×38.7oC=0.231J/goC(3.12.2)(3.12.2)cp=qm×ΔT=134J15.0g×38.7oC=0.231J/goC

Step 3: Think about your result.

The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.

Since most specific heats are known (Table 3.12.13.12.1), they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a 60.0g60.0g of water at 23.52oC23.52oC was cooled by the removal of 813J813J of heat. The change in temperature can be calculated using the specific heat equation.

Since the water was being cooled, the temperature decreases. The final temperature is:

Tf=23.52oC−3.24°C =20.28°C

7.            What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow?

SOLUTION

We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the iron is 73.3°C and the initial temperature is 25.0°C, ΔT is as follows:

ΔT = Tfinal − Tinitial = 73.3°C − 25.0°C = 48.3°C

The mass is given as 150.0 g, and Table 7.3 gives the specific heat of iron as 0.108 cal/g•°C. Substitute the known values into heat = mcΔT and solve for amount of heat:

SOLUTION

The question gives us the heat, the final and initial temperatures, and the mass of the sample. The value of ΔT is as follows:

ΔT = Tfinal − Tinitial = 22.0°C − 97.5°C = −75.5°C

If the sample gives off 71.7 cal, it loses energy (as heat), so the value of heat is written as a negative number, −71.7 cal. Substitute the known values into heat = mcΔT and solve for c:

−71.7 cal = (10.3 g)(c)(−75.5°C)

9.    The specific heat capacity of aluminum is 0.897 J/g°C.   Determine the amount of heat released when 10.5g of aluminum cools from 240°C to 25°C.

                                H  =  m x (sh) x ΔT

                             =   10.5g (0.897 J/g°C) (215) = 2024.98 J

 

10.    The specific heat capacity of aluminum is 0.897 J/g°C.   What mass of aluminum can be heated from 33°C to 99°C using 450J of heat?

                                H  =  m x (sh) x ΔT

                                450 J = m (0.897 J/g°C) (66)

                                7.6g = m