7.1 Internal energy and specific heat capacity — Learn

Prerequisites: work, energy and power from Section 3.3; current, potential difference and electrical power from Section 5.1; SI prefixes from Section 1.1.

By the end of this section you can: explain what internal energy is; explain how temperature and internal energy differ; use $E = mc\Delta\theta$ ; find a specific heat capacity from an experiment; explain what happens to the energy during a change of state; use $E = mL$ ; and read a heating curve.

1. Internal energy

A system is the object, or the group of objects, that we are studying. Its internal energy is the total kinetic energy plus the total potential energy of all the particles in it. The particles are always moving, so they always have kinetic energy. They also pull on one another, so they have potential energy.

The temperature of a system tells you the average random kinetic energy of its particles. It does not tell you the total energy stored.

This difference matters. A full bath of warm water at 40 °C stores far more internal energy than a teaspoon of boiling water at 100 °C. The bath is cooler, but it contains very many more particles.

Heating a system transfers energy to it. Its particles move faster on average, so its internal energy rises and its temperature rises. Doing work on a system, for example by rubbing or by compressing it, also transfers energy and changes its internal energy.

Watch out. Temperature is not the energy stored. A hot spark from a firework is at a very high temperature, but it stores very little internal energy. A bath is at a lower temperature, but it stores far more. Temperature measures the average kinetic energy of one particle. Internal energy is the total over all the particles.

2. Specific heat capacity

The energy needed to warm a system depends on its mass, the material it is made of, and the temperature change. These are linked by

E = mc\Delta\theta

where $E$ is the energy transferred, in joules (J); $m$ is the mass, in kilograms (kg); $c$ is the specific heat capacity, in J kg⁻¹ K⁻¹; and $\Delta\theta$ is the temperature change.

The specific heat capacity of a substance is the energy needed to raise the temperature of 1 kg of it by 1 K. Water has a large specific heat capacity, 4200 J kg⁻¹ K⁻¹. That is why water takes a long time to warm up, and why it carries energy around a central heating system.

A change of 1 °C is the same as a change of 1 K, so $\Delta\theta$ is the same number in either unit.

Worked example 1. A pressure cooker holds 1.50 kg of water at 20 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. Find the energy needed to heat the water to 121 °C.

\begin{aligned} E &= mc\Delta\theta \\ &= 1.50 \times 4200 \times (121 - 20) \\ &= 1.50 \times 4200 \times 101 \\ &= 6.36 \times 10^{5}\,\text{J} \end{aligned}

The water needs $6.36 \times 10^{5}\,\text{J}$ to reach its boiling point. The temperature change is 101 K, which is the same as 101 °C.

3. Changes of state and latent heat

A change of state happens when a substance turns from one state to another. Ice melts to water, water boils to steam, steam condenses back to water, and water freezes to ice.

During a change of state the temperature does not change. The energy supplied is used to break the bonds between the particles, not to speed them up. The potential energy of the particles rises, so the internal energy rises, but their average kinetic energy stays the same.

The specific latent heat of a substance is the energy needed to change the state of 1 kg of it, with no change in temperature:

E = mL

where $E$ is the energy transferred (J), $m$ is the mass (kg), and $L$ is the specific latent heat (J kg⁻¹). There are two kinds to know:

the specific latent heat of fusion, for melting and freezing;
the specific latent heat of vaporisation, for boiling and condensing.

Worked example 2. The specific latent heat of vaporisation of water is $2.26 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ . Find the energy needed to turn 0.050 kg of boiling water into steam at the same temperature.

\begin{aligned} E &= mL \\ &= 0.050 \times 2.26 \times 10^{6} \\ &= 1.13 \times 10^{5}\,\text{J} \end{aligned}

Boiling 0.050 kg of water needs $1.13 \times 10^{5}\,\text{J}$ . Notice that this is a lot of energy for a small mass. Breaking the bonds between water molecules takes much more energy than warming them.

4. Heating and changing state together

Many real processes do both: they warm a substance and then change its state. Work out the energy for each stage, then add the two amounts.

Worked example 3. Find the total energy needed to heat 1.50 kg of water from 20 °C to 121 °C and then turn 0.050 kg of it into steam. Use $c = 4200\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ and $L = 2.26 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ .

\begin{aligned} E_{\text{heat}} &= mc\Delta\theta \\ &= 1.50 \times 4200 \times 101 = 6.36 \times 10^{5}\,\text{J} \\ E_{\text{boil}} &= mL \\ &= 0.050 \times 2.26 \times 10^{6} = 1.13 \times 10^{5}\,\text{J} \\ E_{\text{total}} &= 6.36 \times 10^{5} + 1.13 \times 10^{5} = 7.49 \times 10^{5}\,\text{J} \end{aligned}

The two stages need very different amounts of energy for very different masses. This is the chapter problem, and the Review returns to it.

Worked example 4. A student measures the specific heat capacity of aluminium. An aluminium block of mass 1.0 kg is heated by a 12 V, 2.0 A heater for 300 s. Its temperature rises by 8.0 K. Find the specific heat capacity of the aluminium.

\begin{aligned} E &= IVt \\ &= 2.0 \times 12 \times 300 = 7200\,\text{J} \\ c &= \frac{E}{m\Delta\theta} \\ &= \frac{7200}{1.0 \times 8.0} = 900\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1} \end{aligned}

The specific heat capacity is $900\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ . In the experiment, some energy warms the room and the block loses energy to its surroundings, so the true value is a little lower than this.

Worked example 5. A 700 W heater warms 2.0 kg of water for 300 s. Find the temperature rise. The specific heat capacity of water is $4200\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ .

\begin{aligned} E &= Pt = 700 \times 300 = 2.1 \times 10^{5}\,\text{J} \\ \Delta\theta &= \frac{E}{mc} = \frac{2.1 \times 10^{5}}{2.0 \times 4200} = 25\,\text{K} \end{aligned}

The water warms by 25 K.

5. Reading a heating curve

A graph of temperature against time for a substance heated at a steady rate is a heating curve. It has sloping parts where the substance is warming up, and flat parts where it is changing state. The flat parts are called plateaus.

During a plateau, the energy supplied is changing the state, so the temperature stays the same. The internal energy still rises, because the potential energy of the particles rises.

The length of a plateau shows the latent heat. For water, the boiling plateau is much longer than the melting plateau. The specific latent heat of vaporisation is $2.26 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ , while the specific latent heat of fusion is only $3.34 \times 10^{5}\,\text{J}\,\text{kg}^{-1}$ . The same mass of water needs about seven times as much energy to boil as to melt.

Watch out. A plateau does not mean that the energy has stopped arriving. Energy is still being supplied at a steady rate, and the internal energy is still rising. The temperature is constant because the energy is changing the arrangement of the particles, not their average speed. If you read the plateau as "no energy", every later calculation will be too small.

6. Try it: the heating curve

The figure draws the heating curve of water. Two sliders set the mass of water and the power of the heater.

Predict first. Do not move the sliders yet. Write down your answers.

The graph has a flat part at 0 °C where ice melts, and a flat part at 100 °C where water boils. Which flat part is longer?
If you double the mass of water, what happens to the length of each flat part?
If you double the power of the heater, what happens to the total time?

Test. Set the mass to 1.00 kg and the power to 500 W. Watch the two flat parts and read the times. Then change one slider at a time and check each prediction.

Explain. Why does the graph stay flat while the water is boiling, even though the heater is still supplying energy? Write two or three sentences. Use the idea that the energy is breaking bonds between the particles, so the potential energy rises while the average kinetic energy stays the same.

Progress check

Return to the pressure cooker. It holds 1.50 kg of water at 20 °C, and the water boils at 121 °C.

Find the temperature change as the water is heated to boiling.
Find the energy needed to heat the water to boiling. Use $c = 4200\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ .
Find the energy needed to turn 0.050 kg of the boiling water into steam. Use $L = 2.26 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ .
Explain why the temperature stays at 121 °C while the water boils.

If question 1 goes wrong, subtract 20 from 121. If question 2 goes wrong, use $E = mc\Delta\theta$ with $\Delta\theta = 101\,\text{K}$ . If question 3 goes wrong, use $E = mL$ . If question 4 goes wrong, say what the energy is doing to the bonds between the particles.

You should now be able to

explain what internal energy is;
explain how temperature differs from internal energy;
use $E = mc\Delta\theta$ to find an energy, a mass, a specific heat capacity or a temperature change;
find a specific heat capacity from an electrical heating experiment;
explain what happens to the particles during a change of state;
use $E = mL$ for melting, boiling, freezing and condensing;
work out the total energy when a substance warms and then changes state;
read a heating curve and explain its flat parts.