Thermal Properties and Temperature Worked ExamplesIGCSE Physics 0625

(a) A flat section means energy is going in but the temperature is not rising — a change of state. The first flat section is at the melting temperature of water.

(b) After all the ice has melted (point C), the liquid water warms until it reaches its boiling temperature (point D).

(c) In section DE the water is boiling at $100\,^{\circ}\text{C}$. The marking points for the explanation are:

• the energy supplied is used to change the state (liquid to gas), not to raise the temperature;
• it does work separating the particles against the forces between them;
• the average kinetic energy of the particles does not change, so the temperature stays constant.

(a) Link the temperature to the length of the metal:

• in summer the cables are hotter, so the metal expands and the cables become longer;
• a longer cable between the same two pylons hangs lower (sags more).

(b) On a hot day each section of concrete expands. With no gaps, the sections would push against each other, producing very large forces that could crack or buckle the bridge.

(c) The hot water warms the metal lid, which expands more than the glass rim (the glass is a poor conductor and warms only slowly). The lid becomes slightly larger and looser, so it turns more easily.

When any substance is heated, its particles gain kinetic energy and move faster, so their average separation increases. How much the separation can increase depends on the forces between the particles.

In a solid, strong forces hold the particles close to fixed positions, so they only vibrate a little further apart — very small expansion. In a liquid the forces are weaker, and in a gas they are very weak, so gas particles can move much further apart — the largest expansion.

Check the distractors. A is a classic misconception: particles never change size. C is false — a gas has far fewer particles in the same volume. D is wrong because the particles of a solid vibrate, and vibrate more strongly as the temperature rises; solids do expand, just by less.

(a) Both "per unit" ideas are needed for full credit: specific heat capacity is the energy required per unit mass per unit temperature increase.

(b) Convert to SI units first: grams to kilograms and kilojoules to joules. Then find the temperature rise.

Write the equation, then substitute.

Evaluate and give the unit.

Rearrange $\Delta E = mc\Delta\theta$ to make the temperature rise the subject.

$\Delta E$ and $m$ are the same for both liquids, so $\Delta\theta$ is inversely proportional to $c$. Doubling $c$ halves the temperature rise.

A reverses the relationship. C ignores the specific heat capacity altogether. D is the opposite of the definition: a larger specific heat capacity means more energy is needed per kilogram per degree, so X needs twice as much, not half.

(a) Read the energy at each temperature from the line for $1.0\text{ kg}$, then subtract.

The graph is for $1.0\text{ kg}$. The energy needed is proportional to the mass, so scale by $0.30$.

(b) Use the whole line for the most accurate value: $160\text{ kJ}$ raises $1.0\text{ kg}$ by $100 - 20 = 80\,^{\circ}\text{C}$.

(c) Power is energy per unit time, so $t = E / P$. It is a minimum because in practice some energy is lost to the surroundings.

(a) The ammeter, voltmeter and thermometer are already shown. The student still needs to measure the mass of the block and the heating time: a balance and a stop-watch.

(b) Find the power of the heater, then the energy supplied in $5.0$ minutes $= 300\text{ s}$.

Rearrange $\Delta E = mc\Delta\theta$ for $c$ and substitute. The temperature rise is $33.0 - 18.0 = 15.0\,^{\circ}\text{C}$.

(c) Marking points:

• some of the energy supplied is lost to the surroundings (or heats the heater and thermometer) instead of the block;
• so the temperature rise is smaller than it would be if all $14\,400\text{ J}$ went into the block;
• a smaller $\Delta\theta$ in the denominator gives a larger calculated $c$.

Improvement: add more (thicker) insulation or a lid of insulation on top, or put a little oil in the thermometer hole for better thermal contact.

(a) Marking points for the method:

• measure the mass of water using a balance, in an insulated container (e.g. a polystyrene cup with a lid);
• measure the starting temperature with a thermometer;
• record the joulemeter reading, switch on the heater, and stir the water;
• after the temperature has risen by about $20\,^{\circ}\text{C}$, switch off, keep stirring and record the highest temperature and the final joulemeter reading;
• calculate $c = \Delta E / (m\,\Delta\theta)$.

(b) The energy supplied is the change in the joulemeter reading.

Convert the mass to kilograms and find the temperature rise.

Substitute into the equation.

(a) A fall in temperature means a fall in internal energy: energy has been transferred from the drink to the cooler surroundings.

(b) Marking points:

• the particles (molecules) of the drink move more slowly on average;
• the average kinetic energy of the particles is lower.

(c) $-273\,^{\circ}\text{C}$ is absolute zero, the lowest possible temperature. At absolute zero the particles have the least possible kinetic energy, so no more energy can be removed to lower the temperature further.

(a) Each small division is $1\,^{\circ}\text{C}$. Read the top of each liquid column with your eye level with it, to avoid parallax.

(b) Marking points:

• water evaporates from the wet cloth;
• the more energetic (faster) water molecules escape from the surface;
• so the average kinetic energy of the molecules left behind falls — the remaining water is cooler;
• energy is then transferred from the thermometer bulb to the cooler water, so the reading of Q falls.

(c) The reading of Q rises (towards room temperature). Without air movement, water vapour collects near the cloth and fewer escaping molecules are carried away, so the rate of evaporation decreases and there is less cooling.

(d) Any two of:

• evaporation happens at any temperature; boiling happens only at the boiling point;
• evaporation takes place only at the surface; boiling takes place throughout the liquid (bubbles form inside it);
• evaporation cools the liquid; during boiling the temperature stays constant.

(a) Find the energy gained by the water. Its temperature rises by $25.0 - 20.0 = 5.0\,^{\circ}\text{C}$.

Assume no energy is lost: the energy lost by the metal equals the energy gained by the water. The metal cools from $100\,^{\circ}\text{C}$ to $25.0\,^{\circ}\text{C}$, a fall of $75.0\,^{\circ}\text{C}$.

Rearrange for $c$.

(b) Both errors make the calculated value too small:

• some energy from the metal is lost to the surroundings (and to the cup) instead of to the water, so the water's temperature rise is smaller than it should be;
• the metal cools a little while it is being moved, so its temperature when it enters the water is below $100\,^{\circ}\text{C}$ and its real temperature fall is less than $75.0\,^{\circ}\text{C}$.

(a) Both liquids receive energy at the same rate and have the same mass. B's temperature rises more slowly, so B needs more energy per kilogram per degree: B has the larger specific heat capacity.

(b) Find each gradient (rate of temperature rise) from the whole line.

In one second the heater supplies $50\text{ J}$, so $50 = m \times c \times (\text{rise per second})$. Rearrange for $c$.

Check B another way: in $300\text{ s}$ the heater supplies $50 \times 300 = 15\,000\text{ J}$, and $15\,000 / (0.25 \times 2000) = 30\,^{\circ}\text{C}$ — matching the rise from $20$ to $50\,^{\circ}\text{C}$.

(c) B must rise by $60\,^{\circ}\text{C}$ at $0.10\,^{\circ}\text{C/s}$.

(d) The line would curve, becoming less steep as the temperature rises. The hotter the liquid is compared with the room, the faster it loses energy to the surroundings, so less of each second's $50\text{ J}$ goes into raising its temperature.