Chapter 8: Review

The chapter problem, solved

The weather station stores its water as a block of ice. The block has a volume of $2.5 \times 10^{-3}\,\text{m}^3$ and the density of ice is $920\,\text{kg/m}^3$ .

1. The mass of the ice.

\begin{aligned} m &= \rho V \\ &= 920 \times 2.5 \times 10^{-3} \\ &= 2.3\,\text{kg} \end{aligned}

2. The energy needed to melt the ice. The specific latent heat of fusion of ice is $3.3 \times 10^5\,\text{J/kg}$ .

\begin{aligned} E &= mL \\ &= 2.3 \times 3.3 \times 10^5 \\ &= 759\,000\,\text{J} \\ &= 7.6 \times 10^5\,\text{J} \end{aligned}

3. The energy needed to warm the meltwater. The water is warmed from $0\,^\circ\text{C}$ to $45\,^\circ\text{C}$ , so $\Delta\theta = 45\,^\circ\text{C}$ . The specific heat capacity of water is $4200\,\text{J/kg}\,^\circ\text{C}$ .

\begin{aligned} \Delta E &= mc\Delta\theta \\ &= 2.3 \times 4200 \times 45 \\ &= 434\,700\,\text{J} \\ &= 4.3 \times 10^5\,\text{J} \end{aligned}

4. The total energy. Add the two energies.

759\,000 + 434\,700 = 1\,193\,700\,\text{J} = 1.2 \times 10^6\,\text{J}

Melting the ice needs more energy than warming the melted water. This is the prediction from the chapter opener: breaking the forces between the particles takes more energy than raising their average kinetic energy through $45\,^\circ\text{C}$ .

5. The weather balloon. The balloon holds $2.0\,\text{m}^3$ of gas at $100\,\text{kPa}$ in the valley. High above, the pressure is $25\,\text{kPa}$ and the temperature of the gas is the same. The mass of gas does not change, so $p_1V_1 = p_2V_2$ .

\begin{aligned} V_2 &= \frac{p_1V_1}{p_2} \\ &= \frac{100 \times 2.0}{25} \\ &= 8.0\,\text{m}^3 \end{aligned}

The balloon expands to $8.0\,\text{m}^3$ . The pressure outside has fallen to a quarter of its starting value, so the volume is four times larger. The particles inside hit the balloon wall less often, so the gas pushes the balloon outwards until the pressure inside matches the pressure outside.

6. The bicycle pump. The station also uses a bicycle pump. Pushing the handle does work on the trapped air. The work done increases the internal energy of the air, so the particles move faster and the air becomes warmer.

Mixed practice — choose the method first

For each question, say which relationship you will use before you calculate. Keep your units with your numbers, and watch for the questions that mix a change of temperature with a change of state.

A block has a mass of 3.6 kg and a volume of 4.0 x 10^-3 m^3. Calculate its density.
Calculate the energy needed to warm 1.5 kg of water by 25 °C. The specific heat capacity of water is 4200 J/kg°C.
Calculate the energy needed to melt 0.40 kg of ice at 0 °C. The specific latent heat of fusion of ice is 3.3 x 10^5 J/kg. This is a change of state, so choose the relationship carefully.
A fixed mass of gas at 100 kPa fills 1.5 m^3. It is compressed at constant temperature to 0.50 m^3. Calculate the new pressure.
A gas exerts a pressure of 150 kPa on a surface of area 0.020 m^2. Calculate the force on the surface. This uses pressure from Section 2.4.
A device is supplied with 800 J of energy and transfers 600 J usefully. Calculate its efficiency. This uses efficiency from Chapter 1.
A heater has a power of 1200 W and runs for 3.0 minutes. Calculate the energy it transfers. This uses power from Chapter 1.
A force of 80 N moves an object 6.0 m along the line of the force. Calculate the work done. This uses work done from Chapter 2.
Explain what happens to the energy supplied to a substance while it is changing state, and what happens to the temperature.
H A force of 25 N pushes a piston through 0.20 m. Calculate the work done on the gas. Explain what happens to the internal energy of the gas. This is Higher tier only.
Say which relationships you will use, then calculate the total energy needed to warm 0.50 kg of water from 20 °C to 100 °C and then boil it all away. Use c = 4200 J/kg°C and L = 2.3 x 10^6 J/kg.
A material has a density of 900 kg/m^3. The density of water is 1000 kg/m^3. State whether the material floats on water, and explain your answer using the particle model.
A fixed mass of gas at 120 kPa fills 0.80 m^3 at a constant temperature. Say which relationship you will use, then calculate the volume when the pressure is 60 kPa.

Answers and concise solutions

1. Density of the block

Answer: $\rho = \dfrac{m}{V} = \dfrac{3.6}{4.0 \times 10^{-3}} = 900\,\text{kg/m}^3$ .

2. Energy transferred to the water

Answer: $\Delta E = mc\Delta\theta = 1.5 \times 4200 \times 25 = 157\,500\,\text{J} = 1.6 \times 10^5\,\text{J}$ .

3. Energy needed to melt the ice

Answer: $E = mL = 0.40 \times 3.3 \times 10^5 = 132\,000\,\text{J} = 1.3 \times 10^5\,\text{J}$ .

4. New pressure of the gas

Answer: $p_1V_1 = p_2V_2$ , so $p_2 = \dfrac{100 \times 1.5}{0.50} = 300\,\text{kPa}$ .

5. Force on the surface

Answer: $F = pA = 150\,000 \times 0.020 = 3000\,\text{N}$ .

6. Efficiency of the device

Answer: efficiency $= \dfrac{600}{800} = 0.75$ , which is 75%.

7. Energy transferred by the heater

Answer: $E = Pt$ , with $3.0\,\text{minutes} = 180\,\text{s}$ : $E = 1200 \times 180 = 216\,000\,\text{J} = 2.2 \times 10^5\,\text{J}$ .

8. Work done on the object

Answer: $W = Fs = 80 \times 6.0 = 480\,\text{J}$ .

9. Energy during a change of state

The energy supplied increases the internal energy by breaking the forces between the particles, so their potential energy increases.
The average kinetic energy of the particles does not change, so the temperature stays constant.

10. Work done on the gas and its effect

Answer: $W = Fs = 25 \times 0.20 = 5.0\,\text{J}$ . The work done increases the internal energy of the gas, so the particles move faster and the temperature can rise.

11. Total energy for warming and boiling

Answer: Warming uses $\Delta E = mc\Delta\theta = 0.50 \times 4200 \times 80 = 168\,000\,\text{J}$ . Boiling uses $E = mL = 0.50 \times 2.3 \times 10^6 = 1\,150\,000\,\text{J}$ . Total: $1\,318\,000\,\text{J} = 1.3 \times 10^6\,\text{J}$ .

12. Whether the material floats

The material floats, because its density is less than the density of water.
The material has fewer particles, or less mass, in each unit of volume than water does, so it is less dense and floats.

13. New volume of the gas

Answer: Use $p_1V_1 = p_2V_2$ : $V_2 = \dfrac{120 \times 0.80}{60} = 1.6\,\text{m}^3$ .

What you can now do

use the particle model to describe solids, liquids and gases;
calculate and measure density, and explain why the states have different densities;
describe the changes of state and explain why mass is conserved;
describe internal energy and explain what heating does to the particles;
use $\Delta E = mc\Delta\theta$ and $E = mL$ ;
tell specific heat capacity and specific latent heat apart, and read a heating or cooling graph;
explain gas pressure using collisions with the walls;
use $pV = \text{constant}$ for a fixed mass of gas at constant temperature;
explain how doing work on a gas raises its temperature (Higher tier).

Your relevance reflection

Return to the prompt from the chapter opener: where in your own week does a substance change state, and where does something warm up without changing state? Write one more example of each now that you have finished the chapter. Has the chapter changed what you notice about a kettle, a freezer or the weather?