8.2 Internal energy, energy transfers and particle motions — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Section 8.1 and Chapter 1.

A block has a mass of $1.2\,\text{kg}$ and a volume of $1.5 \times 10^{-3}\,\text{m}^3$ . Calculate its density.
A heater transfers $30\,000\,\text{J}$ of energy in $20\,\text{s}$ . Calculate its power.

Answers

$\rho = \dfrac{m}{V} = \dfrac{1.2}{1.5 \times 10^{-3}} = 800\,\text{kg/m}^3$ .
$P = \dfrac{E}{t} = \dfrac{30\,000}{20} = 1500\,\text{W}$ .

If question 1 went wrong, look again at Section 8.1: divide the mass by the volume. If question 2 went wrong, remember that power is the rate of energy transfer, so divide the energy by the time.

Practise with help

1. A block of copper of mass $0.50\,\text{kg}$ is heated through $80\,^\circ\text{C}$ . The specific heat capacity of copper is $390\,\text{J/kg}\,^\circ\text{C}$ . Calculate the energy transferred to the copper.

Hint

The temperature changes, so use the equation with specific heat capacity. You know the mass, the specific heat capacity and the temperature change.

Next step

Write $\Delta E = mc\Delta\theta$ and put in $m = 0.50$ , $c = 390$ and $\Delta\theta = 80$ .

Full working

\begin{aligned} \Delta E &= mc\Delta\theta \\ &= 0.50 \times 390 \times 80 \\ &= 15\,600\,\text{J} \\ &= 1.56 \times 10^4\,\text{J} \end{aligned}

2. A liquid of mass $1.5\,\text{kg}$ gains $13\,500\,\text{J}$ when its temperature rises by $20\,^\circ\text{C}$ . Calculate the specific heat capacity of the liquid.

Hint

Start from $\Delta E = mc\Delta\theta$ and make $c$ the subject.

Next step

Write $c = \dfrac{\Delta E}{m\Delta\theta}$ , then divide $13\,500$ by $1.5 \times 20$ .

Full working

\begin{aligned} c &= \frac{\Delta E}{m\Delta\theta} \\ &= \frac{13\,500}{1.5 \times 20} \\ &= 450\,\text{J/kg}\,^\circ\text{C} \end{aligned}

3. Calculate the energy needed to melt $0.80\,\text{kg}$ of ice at $0\,^\circ\text{C}$ . The specific latent heat of fusion of ice is $3.3 \times 10^5\,\text{J/kg}$ .

Hint

The ice is at its melting point, so the temperature does not change. Use the latent heat equation.

Next step

Write $E = mL$ and multiply $0.80\,\text{kg}$ by $3.3 \times 10^5\,\text{J/kg}$ .

Full working

\begin{aligned} E &= mL \\ &= 0.80 \times 3.3 \times 10^5 \\ &= 264\,000\,\text{J} \\ &= 2.64 \times 10^5\,\text{J} \end{aligned}

4. Calculate the energy needed to boil away $0.050\,\text{kg}$ of water at $100\,^\circ\text{C}$ . The specific latent heat of vaporisation of water is $2.3 \times 10^6\,\text{J/kg}$ .

Hint

Boiling is a change of state, so this is a latent heat question. The specific latent heat of vaporisation is much larger than the value for fusion.

Next step

Write $E = mL$ and multiply $0.050\,\text{kg}$ by $2.3 \times 10^6\,\text{J/kg}$ .

Full working

\begin{aligned} E &= mL \\ &= 0.050 \times 2.3 \times 10^6 \\ &= 115\,000\,\text{J} \\ &= 1.15 \times 10^5\,\text{J} \end{aligned}

5. Two students heat $1.0\,\text{kg}$ of ice at $0\,^\circ\text{C}$ . One student melts the ice completely. The other student warms $1.0\,\text{kg}$ of water from $0\,^\circ\text{C}$ to $45\,^\circ\text{C}$ . Which student transfers more energy? Show your working.

Hint

One change is a change of state, so it uses latent heat. The other is a temperature change, so it uses specific heat capacity. Calculate both energies.

Next step

For melting, use $E = mL$ with $L = 3.3 \times 10^5\,\text{J/kg}$ . For warming, use $\Delta E = mc\Delta\theta$ with $c = 4200\,\text{J/kg}\,^\circ\text{C}$ .

Full working

Melting:

E = mL = 1.0 \times 3.3 \times 10^5 = 3.3 \times 10^5\,\text{J}

Warming:

\Delta E = mc\Delta\theta = 1.0 \times 4200 \times 45 = 1.89 \times 10^5\,\text{J}

Melting needs more energy, $3.3 \times 10^5\,\text{J}$ compared with $1.89 \times 10^5\,\text{J}$ .

6. A heating graph for a pure substance has a flat part. Explain what is happening to the particles during this flat part, and what happens to the temperature.

Hint

A flat part means the energy supplied is not raising the temperature. Think about what else the energy can do.

Next step

Say that the substance is changing state, then say what the energy does to the forces between the particles.

Full working

On the flat part the substance is changing state. The energy supplied does work against the forces between the particles, so the potential energy of the particles increases and their spacing changes. The average kinetic energy of the particles does not change, so the temperature stays constant.

This answer is a model answer for you to compare with your own wording.

7. Say which relationships you will use, then calculate the total energy needed to warm $0.20\,\text{kg}$ of water from $20\,^\circ\text{C}$ to $100\,^\circ\text{C}$ and then boil it all away. Use $c = 4200\,\text{J/kg}\,^\circ\text{C}$ and $L = 2.3 \times 10^6\,\text{J/kg}$ .

Hint

There are two stages. One warms the water and changes the temperature. The other boils the water and changes the state.

Next step

For the warming stage use $\Delta E = mc\Delta\theta$ with $\Delta\theta = 80\,^\circ\text{C}$ . For the boiling stage use $E = mL$ . Add the two energies.

Full working

Warming from $20\,^\circ\text{C}$ to $100\,^\circ\text{C}$ :

\Delta E = mc\Delta\theta = 0.20 \times 4200 \times 80 = 67\,200\,\text{J}

Boiling at $100\,^\circ\text{C}$ :

E = mL = 0.20 \times 2.3 \times 10^6 = 460\,000\,\text{J}

Total energy:

67\,200 + 460\,000 = 527\,200\,\text{J} = 5.3 \times 10^5\,\text{J}

The total is $5.3 \times 10^5\,\text{J}$ to two significant figures.