Transfer of Thermal Energy Worked ExamplesIGCSE Physics 0625

(a) Air is a fluid, so the transfer through the room is by convection.

(b) Marking points:

• the cold air from the unit is denser than the warmer air around it;
• so the cold air sinks towards the floor;
• warmer, less dense air is pushed up and moves towards the unit, where it is cooled in turn;
• this sets up a convection current that circulates through the whole room, including Y.

(c) Cold air released near the floor is already at the bottom of the room, so it has nowhere to sink to. It forms a cold layer at floor level, and the warm air near the ceiling (at X) is not drawn down, so no convection current forms.

(a) The better the conductor, the faster energy reaches the far end, so the shorter the time.

(b) Any two quantities that would change the time if they varied: the length of the rods, the diameter (thickness) of the rods, the depth each rod is placed in the water, and the temperature of the water.

(c) Glass: its far end had not warmed by $5\,^{\circ}\text{C}$ even after $600\text{ s}$, so it is a bad thermal conductor (a thermal insulator).

(a) In all solids, the particles at the hot end vibrate more strongly and pass energy on to their neighbours: conduction by lattice vibrations. This is the only mechanism in the plastic, and it is slow.

Copper is a metal, so it also has free (delocalised) electrons. At the hot end these electrons gain kinetic energy, move quickly through the metal and collide with ions (lattice particles) further along, transferring energy. This is much faster than lattice vibrations alone.

(b) In a gas the particles are far apart, so they collide with each other much less often. Energy can only be passed on when particles collide, so it is transferred very slowly.

The rate of emission is less for a surface that is light-coloured and shiny, has a smaller surface area and is at a lower temperature.

D has all three: shiny white (a poor emitter), small area and the lower temperature. So D emits at the smallest rate.

A is shiny but large and hotter. B is small and cooler but matt black, which is a good emitter. C has the greatest rate of emission of the four — choosing C means the question was misread as asking for the largest rate.

(a) Average rate of cooling $=$ temperature fall $\div$ time taken.

Repeat for B, keeping to 2 significant figures.

(b) The matt black can cools faster. Evidence: over $600\text{ s}$, A fell by $24.0\,^{\circ}\text{C}$ but B fell by only $14.0\,^{\circ}\text{C}$; and in each interval A's rate is larger ($0.045$ vs $0.025\,^{\circ}\text{C/s}$). A matt black surface is a better emitter of infrared radiation.

(c) Each can cools more slowly in the second interval, when it is cooler ($0.045 \to 0.035\,^{\circ}\text{C/s}$ for A). So the rate of cooling depends on temperature: if one can started hotter it would cool faster anyway, and the comparison of the surfaces would not be fair.

(d) A lid on each can reduces energy loss from the water surface by convection and evaporation, so that the difference between the cans is due mainly to radiation from their surfaces. Both cans must be treated the same.

(a) Thermal radiation is infrared radiation.

(b) Marking points:

• a polished (shiny) surface is a good reflector of infrared radiation;
• so radiation emitted upwards by the element is reflected down towards the food instead of being absorbed by the hood.

(c) At first the food absorbs energy at a greater rate than it transfers energy away (by emission of radiation and to the air), so its internal energy and temperature rise.

As the food gets hotter it transfers energy away faster. When the rate of energy transfer away equals the rate at which it receives energy, the temperature stays constant.

For an object to stay at a constant temperature, it must transfer energy away at the same rate as it receives energy. So B.

A is wrong: the Sun still shines on the rock, so it keeps absorbing. C is wrong: every object emits infrared radiation all the time — the hotter it is, the more it emits. D describes the morning, when the temperature was rising.

(a) Follow the energy step by step:

• the hot coolant flows through the tubes, carrying energy from the engine (convection in the liquid);
• energy passes through the metal walls of the tubes and into the fins by conduction;
• the air next to the fins is warmed, becomes less dense and is carried away by convection — forced by the fan and the motion of the car;
• a small amount of energy is also emitted from the surfaces by radiation (infrared).

(b) Many thin fins give a very large surface area in contact with the air, so energy is transferred to the air (and radiated) at a much greater rate.

(c) Metal is a good thermal conductor (it has free electrons), so energy is conducted quickly from the tubes to the whole of each fin.

(d) A matt black surface is a good emitter of infrared radiation, so the radiator loses more energy by radiation.

(a) Marking points:

• a white surface is a good reflector (poor absorber) of infrared radiation from the Sun;
• so the walls absorb energy at a lower rate and less energy is transferred into the house.

(b) Marking points:

• a matt black surface is a good absorber of infrared radiation;
• so the panel absorbs energy from the Sun at a high rate and transfers it (by conduction) to the water flowing through it.

(a) Marking points:

• in a gas the particles are far apart;
• they collide infrequently, so energy is passed from particle to particle very slowly;
• air is therefore a much worse conductor than glass.

(b) In a wide gap, the air next to the warm pane would rise and the air next to the cold pane would sink, setting up a convection current that carries energy across. A narrow gap restricts the movement of the air, so convection is greatly reduced.

(c) The warm room and inner pane emit infrared radiation. The shiny coating reflects much of this infrared back into the room and is a poor emitter itself, so less energy is lost by radiation.

Apparatus: a thermometer and a stop-watch (plus a measuring cylinder for the water).

Method: pour a measured volume of hot water into the beaker with no insulation; when the thermometer reads a chosen starting temperature (e.g. $80\,^{\circ}\text{C}$), start the stop-watch; record the temperature after a fixed time (e.g. $10$ minutes). Repeat with $1$, $2$, $3$ and $4$ layers of the insulating material.

Control variables: the volume of water, the starting temperature, the same type and size of beaker, the room temperature (away from draughts), and the same cooling time. (These are quantities — the thermometer or the beaker itself is not a variable.)

Table: the heading for each column gives the quantity and its unit.

Conclusion: calculate the temperature fall (or the rate of cooling $=$ fall $\div$ time) for each number of layers. Plot a graph of temperature fall against number of layers. If the fall decreases as the number of layers increases, more insulation reduces the rate of cooling.

(a) Average rate $=$ temperature fall $\div$ time interval.

(b) The rate of cooling decreases as time goes on (from $0.0867$ to $0.0500\,^{\circ}\text{C/s}$; the curve becomes less steep). As the water cools, the difference between its temperature and the room's gets smaller, so it transfers energy to the surroundings more slowly.

(c) The rate will be lower than $0.0867\,^{\circ}\text{C/s}$. In the first experiment, when the water was cooling from about $65\,^{\circ}\text{C}$ (from $180$ to $270\text{ s}$), the rate was only about $0.066\,^{\circ}\text{C/s}$.

(d) The results show that hotter water cools faster. A beaker that started at a higher temperature would have a larger fall in the same time even if its insulation were the same, so the difference in total fall would not be caused only by the factor being tested.