Back to the on-screen lesson · Print

Heat capacities, latent heat and enthalpy

Why $C_P = C_V + nR$, enthalpy $H = U + PV$ as the heat at constant pressure, latent heats of melting and boiling, and the difference between the energy and the enthalpy of a phase change.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to find heat capacities at constant pressure and $\gamma$ for ideal gases, use enthalpy for the heat at constant pressure, price every stage of a heating curve including phase changes, and separate the energy from the work in boiling.

2. What you already have

You can find the heat capacity at constant volume by counting degrees of freedom, and you can compute the work a gas does as it expands. This lesson asks what happens when both happen at once — heating at constant pressure — and what heat does when it goes into a substance without changing its temperature at all.

3. Words for this lesson

4. Heat at constant pressure is a change of enthalpy

Heat a gas in a rigid tank and every joule raises its thermal energy. Heat it in a cylinder under a free piston and the gas expands as it warms, pushing the piston and the atmosphere back. Part of the heat pays for that work, so more heat is needed per kelvin:

$$C_P = C_V + \frac{P\,dV}{dT} = C_V + nR \quad \text{(ideal gas)}.$$

For a monatomic gas that is $\tfrac{5}{2}R$ per mole, for a diatomic gas $\tfrac{7}{2}R$. Their ratio $\gamma = C_P/C_V$ is the exponent of the adiabat.

Most of chemistry and everyday heating happens at constant pressure — open to the atmosphere — and the work term is a nuisance to keep writing. So define the enthalpy

$$H = U + PV.$$

At constant pressure $\Delta H = \Delta U + P\Delta V$, and the first law $\Delta U = Q - P\Delta V$ turns this into

$$Q = \Delta H \quad \text{(constant pressure)}.$$

Enthalpy is the energy you would need to create the system from nothing and push the atmosphere out of the way to make room for it.

Heat does not always change a temperature. At a phase change — ice melting at $0$ °C, water boiling at $100$ °C — the energy supplied separates the molecules instead of speeding them up, and the temperature holds still until the change is complete. The heat per kilogram is the latent heat, $Q = mL$, and tables list it as an enthalpy because phase changes are measured at constant pressure.

5. The method, step by step, and how to check it

A heating problem is a budget, and the method is to list what the heat pays for.

1. Draw the heating curve if the substance changes phase: temperature against energy supplied, with sloping stretches where one phase warms and flat stretches where the phase changes.
2. Price each sloping stretch with $Q = mc\Delta T$ or $Q = nC\Delta T$, using the heat capacity of the phase actually present — ice, water and steam have different specific heats ($2.09$, $4.19$ and about $2.01$ kJ/(kg K)).
3. Price each flat stretch with $Q = mL$.
4. For a gas at constant pressure, use $C_P$, and if asked, split the heat into $\Delta U = C_V\Delta T$ and work $P\Delta V = nR\Delta T$.
5. Add the stretches; energy is conserved.

Checks. At constant pressure a gas always needs more heat than at constant volume, by exactly $nR$ per kelvin for an ideal gas. Boiling takes far more energy than melting, since separating molecules completely costs more than loosening them: for water, $L_v$ is almost seven times $L_f$. And the temperature on a flat stretch is fixed by the pressure, not by how fast you heat: turning up the burner under boiling water boils it away faster but does not make it hotter.

6. Energy against enthalpy when a liquid boils

When one mole of water boils at $100$ °C and atmospheric pressure, $40.7$ kJ of heat goes in: that is the enthalpy of vaporization. How much of it is the energy needed to pull the molecules apart, and how much is work?

The vapor occupies $V = nRT/P = (1)(8.31)(373)/(1.013 \times 10^{5}) = 0.0306$ m³, about $30$ litres, where the liquid occupied $18$ millilitres. Making room for it at atmospheric pressure costs

$$P\Delta V \approx nRT = 8.31 \times 373 = 3.10\ \text{kJ}.$$

So the internal energy rises by $\Delta U = 40.7 - 3.1 = 37.6$ kJ, and about $7.6$ percent of the heat went into the atmosphere as work. In a vacuum, where there is nothing to push back, boiling would need only the $37.6$ kJ — which is part of why water boils at a lower temperature on a mountain, where the atmosphere pushes less.

The same distinction runs through chemistry. The heat of combustion of methane, $\Delta H = -890$ kJ/mol at constant pressure, differs slightly from the energy change measured in a sealed bomb calorimeter, because the reaction turns three moles of gas into one (with the water condensed), and the atmosphere does work on the shrinking products.

7. Why water is so good at storing heat

Water's specific heat, $4186$ J/(kg K), is one of the largest of any common substance — ten times iron's per kilogram — and its latent heats are enormous. In the language of the last lesson, a water molecule has many ways to store energy: it translates and rotates about three axes, and in the liquid each molecule is held by hydrogen bonds to its neighbors, which stretch and bend and break as the temperature rises. That is why water is the working fluid of steam engines and power plants, the coolant in car engines and the medium of home heating systems.

It also governs climate. The oceans store about a thousand times more heat per degree than the whole atmosphere, so coastal cities have milder summers and winters than inland ones at the same latitude: San Francisco's average temperature varies by about $10$ °F between January and September, while Kansas City's swings by about $50$ °F. And the latent heat carried by evaporating seawater, released when the vapor condenses into rain, is the engine of every hurricane.

8. In the world: why steam burns are worse than boiling-water burns

Hold a hand in the steam above a pot for a moment and it can burn worse than a splash of boiling water at the same $100$ °C. The latent heat explains why.

When $1$ g of water at $100$ °C touches skin and cools to $37$ °C, it gives up $mc\Delta T = 0.001 \times 4186 \times 63 = 264$ J. When $1$ g of steam at $100$ °C touches skin, it first condenses, releasing its latent heat of vaporization, $mL_v = 0.001 \times 2.257 \times 10^{6} = 2257$ J, and then cools from $100$ °C to $37$ °C as liquid, releasing another $264$ J: $2521$ J in all, almost ten times as much.

Steam also condenses preferentially on the coolest surface nearby, which is your skin, and it keeps arriving for as long as the hand stays in the plume. The same numbers make steam an excellent carrier of heat on purpose: in the district steam systems that still heat much of Manhattan, a kilogram of steam piped to a building delivers over two megajoules when it condenses in a radiator, several times what a kilogram of hot water could deliver by cooling the same number of degrees.

9. In the world: sweating and the latent heat of evaporation

A person working hard on a hot day may lose a liter of sweat an hour, and evaporating it is how the body sheds heat when the air is as warm as the skin. At skin temperature the latent heat of vaporization of water is about $2.4$ MJ/kg, slightly higher than at $100$ °C because the molecules have less thermal energy to start with.

One kilogram of sweat evaporating in an hour removes $2.4 \times 10^{6}$ J in $3600$ s, a cooling power of about $670$ W — several times the $100$ W a resting body produces, and enough to carry the heat of hard exercise. The weakness is the word evaporating. Sweat that drips off removes almost nothing, and in humid air, where the vapor pressure of the air is close to that of the sweat, evaporation slows to a crawl. That is why the National Weather Service's heat index combines temperature with humidity: $95$ °F at $60$ percent relative humidity feels like $113$ °F, because the body's main cooling system is failing.

10. Heat is not always a temperature rise

The habit $Q = mc\Delta T$ is learned first and sticks, so it is natural to think that adding heat always raises a temperature. At a phase change it does not: a pot of boiling water stays at $100$ °C however high the flame, and an ice-water mixture stays at $0$ °C until the last of the ice melts. The heat is going into separating molecules, and the thermometer does not see it. Heating curves have flat stretches for exactly this reason.

A second confusion is between enthalpy and energy. The enthalpy of vaporization is the heat measured at constant pressure; it is not the energy the molecules gain. A part of it — about eight percent for water — goes into the work of pushing the atmosphere back. The two coincide only for processes with no change of volume, which is why the distinction hardly matters for melting and matters a great deal for boiling and for reactions that make or consume gas.

11. Heating nitrogen at constant pressure

1. $2.0$ mol of nitrogen is heated at constant pressure by $50$ K. Find its molar heat capacity at constant volume.

   $\dfrac{C_V}{n} = \tfrac{5}{2}R = 20.775\ \text{J/(mol K)}$

   Nitrogen has five active degrees of freedom.

2. Add $R$ to get the molar heat capacity at constant pressure.

   $\dfrac{C_P}{n} = 20.775 + 8.31 = 29.085\ \text{J/(mol K)}$

   Each mole expanding by one kelvin does $R$ joules of work.

3. Find the heat supplied.

   $Q = n \cdot \dfrac{C_P}{n} \cdot \Delta T =2.0 \times 29.085 \times 50 = 2908.5\ \text{J}$

   At constant pressure $Q = \Delta H = C_P\Delta T$.

4. Find the rise in thermal energy.

   $\Delta U = 2.0 \times 20.775 \times 50 = 2077.5\ \text{J}$

   Thermal energy depends only on temperature, so this is the same as it would be at constant volume.

5. Find the work done by the gas as the difference.

   $P\Delta V = Q - \Delta U = 2908.5 - 2077.5 = 831\ \text{J} = nR\Delta T$

   The check, $2.0 \times 8.31 \times 50 = 831$, confirms the budget balances.

12. Ice to steam

1. $0.50$ kg of ice at $0$ °C is to be turned into steam at $100$ °C. Melt the ice.

   $Q_1 = mL_f = 0.50 \times 333 = 166.5\ \text{kJ}$

   The temperature stays at $0$ °C until all the ice has melted.

2. Warm the water to its boiling point.

   $Q_2 = mc\Delta T = 0.50 \times 4.186 \times 100 = 209.3\ \text{kJ}$

   Now the energy speeds the molecules up, and the temperature climbs.

3. Boil the water.

   $Q_3 = mL_v = 0.50 \times 2257 = 1128.5\ \text{kJ}$

   The temperature stays at $100$ °C until all the water has boiled.

4. Add the stages.

   $Q = 166.5 + 209.3 + 1128.5 = 1504.3\ \text{kJ}$

   Energy is conserved, so the stages add.

5. Find the share taken by boiling.

   $\dfrac{1128.5}{1504.3} \approx 0.75$

   Three quarters of the energy goes into the last, flat stretch.

6. Find how long a $2.0$ kW stove burner would take.

   $t = \dfrac{1.504 \times 10^{6}}{2000} \approx 752\ \text{s} \approx 12.5\ \text{min}$

   Ignoring losses to the pan and the air, which in practice are substantial.

13. Energy and enthalpy of boiling water

1. One mole of water boils at $373$ K and $101$ kPa, absorbing $40.7$ kJ. Name the heat.

   $Q = \Delta H = 40.7\ \text{kJ}$

   At constant pressure the heat equals the change in enthalpy.

2. Find the volume of the vapor.

   $V = \dfrac{nRT}{P} = \dfrac{1 \times 8.31 \times 373}{1.01 \times 10^{5}} \approx 0.0307\ \text{m}^3$

   The vapor behaves nearly as an ideal gas.

3. Compare it with the liquid's volume.

   $V_{\text{liquid}} = 18\ \text{mL} = 1.8 \times 10^{-5}\ \text{m}^3 \ll 0.0307\ \text{m}^3$

   So the change in volume is essentially the vapor's volume.

4. Find the work done on the atmosphere.

   $P\Delta V \approx nRT = 8.31 \times 373 = 3100\ \text{J} = 3.10\ \text{kJ}$

   Using $PV = nRT$ avoids needing the pressure at all.

5. Subtract to get the change in internal energy.

   $\Delta U = \Delta H - P\Delta V = 40.7 - 3.1 = 37.6\ \text{kJ}$

   From $H = U + PV$ at constant pressure.

6. Find the fraction that was work.

   $\dfrac{3.1}{40.7} \approx 7.6\%$

   Most of the heat separates the molecules; a small part makes room for the vapor.

7. Express the energy per molecule.

   $\dfrac{37.6 \times 10^{3}}{6.02 \times 10^{23}} \approx 6.2 \times 10^{-20}\ \text{J} \approx 12\,kT$

   Pulling one molecule out of the liquid costs about twelve times its typical thermal energy at $373$ K, which is why only a few molecules escape at a time.

14. Your turn: find $\gamma$ for helium from its heat capacities.

1. Write the molar heat capacity at constant volume.

   $\dfrac{C_V}{n} = \tfrac{3}{2}R$

   Helium has three degrees of freedom.

2. Add $R$ for constant pressure.

   $\dfrac{C_P}{n} = \tfrac{3}{2}R + R = \tfrac{5}{2}R$

   The extra $R$ is the work of expansion.

3. Your turn: work this step out. Its working is at the end of the packet.

   Take the ratio.

15. Guided practice

$1$ mol of helium is heated at constant pressure by $31$ K. What fraction of the heat supplied goes into the work the gas does as it expands?

• $\dfrac{2}{5}$
• $0$, since the gas only gets hotter
• $\dfrac{3}{5}$
• $\dfrac{2}{3}$

16. Guided practice

Complete the worked solution: $8$ mol of argon is heated at constant pressure by $15$ K. Find the rise in its thermal energy, the work it does and the heat supplied. Use $R = 8.31$ J/(mol K).

1. Multiply out $nR\Delta T$.

   $nR\Delta T = 8 \times 8.31 \times 15 =$ _a J

   Every term of this budget is a multiple of it; it is also the work the gas does, $P\Delta V$.

2. Take three halves of it for the thermal energy.

   $\Delta U = \tfrac{3}{2}nR\Delta T =$ _b J

   Argon is monatomic, with three degrees of freedom.

3. Add the work to the thermal energy for the heat.

   $Q = \Delta U + P\Delta V =$ _c J

   The heat pays for the warming and for the expansion.

17. Guided practice

Match each quantity for an ideal gas at room temperature to its value. The values are per mole, for any of the $1$ samples in a lab.

18. Practice

$0.8$ kg of ice at $0$ °C is turned into steam at $100$ °C. Use $L_f = 333$ kJ/kg, $c = 4.186$ kJ/(kg K) for water and $L_v = 2257$ kJ/kg. Fill in the energy for each stage and the total, in kJ.

19. Practice

$8$ mol of nitrogen in a cylinder with a freely sliding piston is heated at constant atmospheric pressure by $36$ K. How much heat is supplied? Use $R = 8.31$ J/(mol K).

Answer: unit: J / MJ / kJ

20. Practice

The molar enthalpy of vaporization of ammonia at its boiling point of $240$ K and atmospheric pressure is $23.3$ kJ/mol. By how much does the internal energy of $2$ mol of it change when it boils? Treat the vapor as an ideal gas and neglect the volume of the liquid. Use $R = 8.31$ J/(mol K).

Answer: unit: J / MJ / kJ

21. Somewhere new

A $2500$ W electric kettle has brought $0.2$ kg of water to the boil and been left on. How long does it take to boil the water away completely, if all its power goes into the water? Use $L_v = 2257$ kJ/kg.

Answer: unit: h / s / min

22. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

23. Test question

$0.6$ kg of ice at $0$ °C is turned into steam at $100$ °C. Use $L_f = 333$ kJ/kg, $c = 4.186$ kJ/(kg K) for water and $L_v = 2257$ kJ/kg. Fill in the energy for each stage and the total, in kJ.

24. What you can do now

You can budget the heat in a process at constant pressure and through a change of phase. Explain to someone why a steam burn is worse than a boiling-water burn.

Working for the steps left to you

14. Your turn: find $\gamma$ for helium from its heat capacities., step 3