8.2 Internal energy, energy transfers and particle motions — Learn

Prerequisites: Section 8.1, especially the particle model and the changes of state; Chapter 1, especially energy stores, energy transfers and power.

By the end of this section you can: describe the internal energy of a system; explain what heating does to the particles; calculate an energy transfer with $\Delta E = mc\Delta\theta$ and with $E = mL$; tell specific heat capacity and specific latent heat apart; and interpret a heating or cooling graph.

1. Internal energy

Every particle in a substance has energy. It has kinetic energy because it is moving or vibrating, and it has potential energy because of the forces between it and the particles around it.

The internal energy of a system is the total kinetic energy and the total potential energy of all the particles in the system.

Internal energy is not the same as temperature. Temperature measures the average kinetic energy of the particles. A small cup of water at $80\,^\circ\text{C}$ is at a higher temperature than a full bath at $40\,^\circ\text{C}$, but the bath holds much more internal energy, because it contains far more particles.

2. Heating a system

Heating is an energy transfer to a system, from a hotter place to a cooler one. Heating increases the internal energy of the system. The extra energy can do one of two things.

• It can raise the temperature. The particles move faster, so their average kinetic energy increases. The temperature rises.
• It can produce a change of state. The energy breaks the forces between the particles, so their potential energy increases. The temperature does not change while this happens.

The size of the temperature rise depends on three things: the mass of the substance heated, the material it is made of, and the energy supplied to it.

Worked example 1. A student heats $2.0\,\text{kg}$ of water from $20\,^\circ\text{C}$ to $50\,^\circ\text{C}$. The specific heat capacity of water is $4200\,\text{J/kg}\,^\circ\text{C}$. Calculate the energy transferred to the water.

The temperature change is

The water gains $2.5 \times 10^5\,\text{J}$.

3. Specific heat capacity

The specific heat capacity of a substance is the energy needed to raise the temperature of one kilogram of the substance by one degree Celsius. Its unit is $\text{J/kg}\,^\circ\text{C}$.

The equation for a temperature change is

where $\Delta E$ is the change in thermal energy in joules, $m$ is the mass in kilograms, $c$ is the specific heat capacity and $\Delta\theta$ is the temperature change in degrees Celsius.

A material with a large specific heat capacity needs a lot of energy for a small temperature rise. Water has a large value, $4200\,\text{J/kg}\,^\circ\text{C}$. That is why water is used to store thermal energy in a central heating system.

Worked example 2. A block of metal of mass $2.0\,\text{kg}$ gains $46\,200\,\text{J}$ when its temperature rises by $55\,^\circ\text{C}$. Calculate the specific heat capacity of the metal.

Make $c$ the subject of $\Delta E = mc\Delta\theta$.

The metal has a specific heat capacity of $420\,\text{J/kg}\,^\circ\text{C}$.

4. Specific latent heat

Energy is also needed to change the state of a substance. The energy needed to change state is called latent heat. During a change of state the energy supplied increases the internal energy, but the temperature does not change.

The specific latent heat of a substance is the energy needed to change the state of one kilogram of the substance with no change in temperature. Its unit is $\text{J/kg}$.

The equation for a change of state is

where $E$ is the energy in joules, $m$ is the mass in kilograms and $L$ is the specific latent heat in $\text{J/kg}$.

There are two kinds of specific latent heat.

• Specific latent heat of fusion is the energy to change from solid to liquid. For ice, $L = 3.3 \times 10^5\,\text{J/kg}$.
• Specific latent heat of vaporisation is the energy to change from liquid to vapour. For water, $L = 2.3 \times 10^6\,\text{J/kg}$.

Worked example 3. Calculate the energy needed to melt $0.50\,\text{kg}$ of ice at $0\,^\circ\text{C}$. The specific latent heat of fusion of ice is $3.3 \times 10^5\,\text{J/kg}$.

To two significant figures the energy is $1.7 \times 10^5\,\text{J}$.

Worked example 4. Calculate the energy needed to boil away $0.20\,\text{kg}$ of water at $100\,^\circ\text{C}$. The specific latent heat of vaporisation of water is $2.3 \times 10^6\,\text{J/kg}$.

5. Heating and cooling graphs

A heating graph shows temperature on the vertical axis and energy or time on the horizontal axis. It has two kinds of part.

• A sloped part, where the temperature is changing. Here the substance is being heated without changing state, and $\Delta E = mc\Delta\theta$ applies.
• A flat part, where the temperature does not change. Here the substance is changing state, and $E = mL$ applies.

A cooling graph is the same shape read from right to left. The flat parts are still the changes of state.

Worked example 5. A heating graph for a pure substance has a flat part at $0\,^\circ\text{C}$ and another at $100\,^\circ\text{C}$. Explain what is happening on each flat part.

A model answer is:

• On the flat part at $0\,^\circ\text{C}$, the substance is melting. The energy supplied breaks the forces between the particles, so the state changes but the temperature stays the same.
• On the flat part at $100\,^\circ\text{C}$, the substance is boiling. The energy supplied separates the particles completely, so the state changes again, and the temperature stays the same.

6. Heating curves and energy

The figure shows a heating graph for $1.0\,\text{kg}$ of ice, starting at $-20\,^\circ\text{C}$ and ending as steam. You can change the power of the heater.

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

1. How many flat parts will the graph have? Where will they be?
2. If the heater has a higher power, what happens to the length of each flat part in time?

Test. Move the power slider. Watch how long each stage takes. Compare a low power with a high power.

Explain. The energy needed to melt the ice is the same whatever the power of the heater. Explain why a higher power makes the flat part shorter in time but does not change the energy needed. Write your answer in two or three sentences.

Watch out. Specific heat capacity and specific latent heat are not the same thing. Specific heat capacity is used when the temperature changes, and its unit is $\text{J/kg}\,^\circ\text{C}$. Specific latent heat is used when the temperature is constant during a change of state, and its unit is $\text{J/kg}$. A flat part of a heating graph means a change of state, so use $E = mL$ there, not $\Delta E = mc\Delta\theta$. Re-check which part of the graph the question is about before you choose an equation.

Progress check

Return to the chapter problem. The station melts the block of ice, which has a mass of $2.3\,\text{kg}$, and then warms the meltwater from $0\,^\circ\text{C}$ to $45\,^\circ\text{C}$.

1. Calculate the energy needed to melt the ice. Use $L = 3.3 \times 10^5\,\text{J/kg}$.
2. Calculate the energy needed to warm the meltwater. Use $c = 4200\,\text{J/kg}\,^\circ\text{C}$.
3. Which of the two answers is larger? Give a reason for your answer.

You should now be able to

• describe internal energy as the total kinetic and potential energy of the particles in a system;
• explain how heating can raise the temperature or change the state;
• relate the temperature of a substance to the average kinetic energy of its particles;
• define specific heat capacity and use $\Delta E = mc\Delta\theta$;
• define specific latent heat and use $E = mL$;
• distinguish between specific heat capacity and specific latent heat;
• interpret a heating or cooling graph, including its flat parts.