Specific heats and state variables

Specific heat of solids and water

Narrated video. Transcript:

Heat an iron tawa and a pot of water on identical burners for one minute. The tawa gets far hotter. Different substances need different amounts of heat for the same rise in temperature.

If heat ΔQ raises the temperature of a body by ΔT, its heat capacity is S equals ΔQ over ΔT. A bigger body of the same material has a bigger heat capacity.

Divide by the mass to get a property of the material alone: the specific heat capacity, s equals ΔQ over m ΔT, in joules per kilogram per kelvin.

Chemists often count moles instead. The molar specific heat capacity is C equals ΔQ over μ ΔT, where μ is the number of moles. Its unit is joules per mole per kelvin.

Now look at a solid. Each atom vibrates about its fixed place in three directions. Kinetic theory shows that each atom then has an average energy of 3 k_B T.

For one mole, that is 3 k_B T times Avogadro’s number, so U equals 3 R T. A solid hardly expands, so almost all the heat goes into U, and C equals ΔU over ΔT, which is 3R, about 24.9 joules per mole per kelvin.

Check the table. Aluminium, copper, silver, tungsten and lead all have molar specific heats close to 25. Carbon, at about 6, is a big exception at room temperature.

Water is special. The old unit of heat, the calorie, is the heat that raises one gram of water from 14.5 to 15.5 °C. One calorie equals 4.186 joules.

So water’s specific heat is about 4186 joules per kilogram per kelvin, higher than almost any common substance. That is why water cools car engines, and why a hot water bag stays warm for so long.

Gases: Cp − Cv = R

Narrated video. Transcript:

For a gas, the heat needed depends on how you heat it. Here are two cylinders, each with one mole of the same ideal gas at the same temperature.

In the first, the piston is pinned so the volume cannot change. No work is done, so all the heat goes into internal energy. The molar specific heat at constant volume is C v equals ΔU over ΔT.

In the second, the piston is free and carries a fixed load, so the pressure stays constant. As the gas warms, it expands and does work p ΔV as well.

So the heat needed at constant pressure is bigger: C p equals ΔU over ΔT plus p times ΔV over ΔT.

For an ideal gas, U depends only on temperature, so ΔU over ΔT is the same C v in both cylinders.

And for one mole, p V equals R T. At constant pressure, p ΔV equals R ΔT. Put these in, and C p equals C v plus R.

So C p minus C v equals R, the gas constant, 8.31 joules per mole per kelvin. This is Mayer’s relation, and it holds for every ideal gas.

Example: heat 2 moles of helium, a monatomic gas, from 300 K to 350 K. At constant volume, with C v equal to 3 by 2 R, you need about 1247 joules. At constant pressure, with C p equal to 5 by 2 R, you need about 2079 joules.

The difference, about 831 joules, is μ R ΔT. It is exactly the work the gas does pushing the piston out.

State variables and the equation of state

Narrated video. Transcript:

Every equilibrium state of a system is fully described by a few state variables: pressure, volume, temperature, mass and composition. Their values do not depend on how the system reached that state.

Now open a valve and let gas rush into an evacuated chamber. During this free expansion, the pressure and temperature are different from place to place. The gas has no single p or T. It is not in equilibrium.

For an ideal gas in equilibrium, the state variables are linked by the equation of state: p V equals μ R T.

So for a fixed amount of gas, only two of p, V and T are independent. Choose any two, and the third is fixed. That is why one point on a P-V diagram is one complete state.

Now slide a partition into the middle of a box of gas in equilibrium. Look at each half. Pressure, temperature and density are the same as before. These are intensive variables.

But the volume, the mass and the internal energy of each half are halved. These are extensive variables. They scale with the size of the system.

A good check: in ΔQ equals ΔU plus p ΔV, every term is extensive. Pressure is intensive, but multiplying it by ΔV, which is extensive, gives an extensive product.

Try it: Cv and Cp in the lab

Interactive simulation. Things to try:

• Isochoric, T₂ = 450 K: the heat goes only into ΔU. The bottom readout shows ΔQ/(μΔT) = 1.50R = Cv for a monatomic gas.
• Now choose Isobaric and set V₂ = 15 L. The gas also ends at 450 K, but needs more heat: ΔQ/(μΔT) = 2.50R = Cp.
• The extra heat (about 1247 J) equals the work W = pΔV, which is RΔT for one mole.
• Switch to Diatomic and repeat: Cv = 2.5R and Cp = 3.5R, but Cp − Cv is still R.

Cut it in half

A box of gas in equilibrium is divided into two equal parts by a thin partition. Which quantity is halved in each part?

1. Pressure
2. Temperature
3. Density
4. Internal energy

Heating at constant pressure

How much heat is needed to raise the temperature of 2 mol of a diatomic ideal gas from 300 K to 400 K at constant pressure? (Cv = 5R/2, R = 8.314 J mol⁻¹ K⁻¹)

1. 1663 J
2. 4157 J
3. 5820 J
4. 2910 J

Summary

• Heat capacity: S = ΔQ / ΔT
• Specific heat capacity: s = ΔQ / (m ΔT)
• Molar specific heat: C = ΔQ / (μ ΔT)
• Solids (Dulong–Petit): C ≈ 3R ≈ 24.9 J mol⁻¹ K⁻¹
• Mayer’s relation (ideal gas): Cp − Cv = R
• Equation of state: pV = μRT

• Water has a very high specific heat, about 4186 J kg⁻¹ K⁻¹; 1 cal = 4.186 J.
• A gas needs more heat at constant pressure than at constant volume, because it also does work pΔV. So Cp > Cv.
• State variables (p, V, T, mass, composition) describe an equilibrium state, whatever path led there. For a fixed amount of ideal gas, only two of p, V, T are independent.
• Intensive variables (p, T, density) do not depend on the size of the system; extensive ones (V, mass, U) do.