7.1 Internal energy and specific heat capacity — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Keep the mass in kilograms, and remember that a temperature change of 1 °C is the same as a change of 1 K.

Warm-up: remember

A heater transfers 3600 J of energy in 120 s. Find its power.
Write 500 g in kilograms.
The temperature of a liquid rises from 18 °C to 45 °C. Find the temperature change in kelvin.

Answers

$P = \dfrac{E}{t} = \dfrac{3600}{120} = 30\,\text{W}$ .
$500\,\text{g} = 0.50\,\text{kg}$ .
$\Delta\theta = 45 - 18 = 27\,\text{K}$ .

If question 1 goes wrong, remember that power is energy divided by time. If question 2 goes wrong, the prefix kilo- means 1000. If question 3 goes wrong, a change of one degree Celsius is a change of one kelvin.

Practise with help

1. A 0.80 kg copper block is heated from 20 °C to 45 °C. The specific heat capacity of copper is 390 J kg⁻¹ K⁻¹. Find the energy supplied.

Hint

The block is warming up and not changing state, so use the specific heat capacity equation.

Next step

Find the temperature change first: $\Delta\theta = 45 - 20 = 25\,\text{K}$ . Then substitute $m = 0.80\,\text{kg}$ and $c = 390\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ into $E = mc\Delta\theta$ .

Full working

\begin{aligned} E &= mc\Delta\theta \\ &= 0.80 \times 390 \times 25 \\ &= 7800\,\text{J} \end{aligned}

The energy supplied is $7800\,\text{J}$ .

2. Find the energy needed to melt 0.20 kg of ice at 0 °C. The specific latent heat of fusion of ice is $3.34 \times 10^{5}\,\text{J}\,\text{kg}^{-1}$ .

Hint

The ice is at its melting point, so the temperature stays at 0 °C. The state changes, so this is a latent heat problem.

Next step

Use $E = mL$ with $m = 0.20\,\text{kg}$ and $L = 3.34 \times 10^{5}\,\text{J}\,\text{kg}^{-1}$ .

Full working

\begin{aligned} E &= mL \\ &= 0.20 \times 3.34 \times 10^{5} \\ &= 6.68 \times 10^{4}\,\text{J} \end{aligned}

Melting the ice needs $6.68 \times 10^{4}\,\text{J}$ .

3. A 2.0 kg block gains 45000 J of energy and its temperature rises by 50 K. Find its specific heat capacity.

Hint

The equation $E = mc\Delta\theta$ contains $c$ . Rearrange it to make $c$ the subject.

Next step

Divide both sides by $m\Delta\theta$ to get $c = \dfrac{E}{m\Delta\theta}$ , then substitute the three values.

Full working

\begin{aligned} c &= \frac{E}{m\Delta\theta} \\ &= \frac{45000}{2.0 \times 50} \\ &= 450\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1} \end{aligned}

The specific heat capacity is $450\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ .

4. A pan heats 0.50 kg of water from 20 °C to 100 °C, then boils 0.10 kg of it into steam. Use $c = 4200\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ and $L = 2.26 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ . Find the total energy supplied.

Hint

There are two stages here: one where the water warms up, and one where it changes state. Work out the energy for each stage.

Next step

For the warming stage use $E_1 = mc\Delta\theta$ with $\Delta\theta = 80\,\text{K}$ . For the boiling stage use $E_2 = mL$ . Add $E_1$ and $E_2$ at the end.

Full working

\begin{aligned} E_1 &= mc\Delta\theta = 0.50 \times 4200 \times 80 = 1.68 \times 10^{5}\,\text{J} \\ E_2 &= mL = 0.10 \times 2.26 \times 10^{6} = 2.26 \times 10^{5}\,\text{J} \\ E_{\text{total}} &= 1.68 \times 10^{5} + 2.26 \times 10^{5} = 3.94 \times 10^{5}\,\text{J} \end{aligned}

The total energy supplied is $3.94 \times 10^{5}\,\text{J}$ .

5. A heater of 6.0 V and 3.0 A runs for 200 s in 1.5 kg of liquid. The temperature of the liquid rises by 4.0 K. Find the specific heat capacity of the liquid.

Hint

The heater supplies electrical energy. Find the energy it supplies first, using the current, the potential difference and the time.

Next step

The energy supplied is $E = IVt = 3.0 \times 6.0 \times 200$ . Then divide by $m\Delta\theta$ to find $c$ .

Full working

\begin{aligned} E &= IVt = 3.0 \times 6.0 \times 200 = 3600\,\text{J} \\ c &= \frac{E}{m\Delta\theta} = \frac{3600}{1.5 \times 4.0} = 600\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1} \end{aligned}

The specific heat capacity is $600\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ .

6. Explain why the temperature of ice does not change while it is melting, even though energy is being supplied.

Hint

Temperature measures the average random kinetic energy of the particles. Think about what the energy is doing to the arrangement of the particles.

Next step

Say that the energy breaks the bonds that hold the particles in the solid. Then say what happens to the potential energy and to the average kinetic energy.

Full working

Energy is supplied, so the internal energy of the ice rises. The energy is used to break the bonds between the particles, so their potential energy increases. The average random kinetic energy of the particles does not change. Temperature measures that average kinetic energy, so the temperature stays at 0 °C until all the ice has melted.