Thermal physics for Cambridge IGCSE Physics

Reading cue. The 300 cm³ and 120 cm³ are the trapped gas columns, not the whole cylinder, and the note fixing the temperature is what allows pressure and volume to be linked at all.

1. aIdentify Identify two quantities that are the same for the trapped gas in both drawings of Fig. 2.1, and state how you know each from the figure.

   recall2 marks

   Check answer · 2 marks

   1. the mass, or number, of gas molecules is the same, because the cylinder is closed at one end and the piston seals the other so no gas escapes (1)
   2. the temperature is the same, as stated in the note below the drawings (1)

2. bDetermine Determine the volume the trapped gas would occupy if the piston were pushed in until the pressure reached 4.0 × 10⁵ Pa at the same temperature.

   routine3 marks

   Check answer · 3 marks

   1. pressure x volume is constant at constant temperature: 1.0 × 10⁵ x 300 = 4.0 × 10⁵ x V (1)
   2. V = 3.0 × 10⁷ / 4.0 × 10⁵ (1)
   3. V = 75 cm³ (1)

3. cExplain Explain, in terms of the molecules of the trapped gas, why the pressure in the lower drawing of Fig. 2.1 is greater than in the upper drawing.

   demanding4 marks

   Check answer · 4 marks

   1. the same number of molecules is now contained in a smaller volume (1)
   2. the temperature is unchanged, so the average speed and average kinetic energy of the molecules are unchanged (1)
   3. each molecule has a shorter distance to travel between the walls, so it strikes them more often (1)
   4. more collisions each second on each unit area of wall gives a greater average force per unit area, and so a greater pressure (1)

4. dSuggest The piston is instead pushed in very quickly from 300 cm³ to 120 cm³, so that the note below Fig. 2.1 no longer applies. Suggest how the pressure reached compares with the value for a slow compression, and explain your answer.

   top of the paper4 marks

   Check answer · 4 marks

   1. the pressure reached is greater than the constant-temperature value (1)
   2. the piston does work on the gas and there is no time for that energy to pass out to the surroundings (1)
   3. the temperature of the gas rises, so the molecules move faster on average (1)
   4. faster molecules strike the walls harder and more often, raising the pressure further; as the gas then cools to room temperature the pressure falls back towards the constant-temperature value (1)

Reading cue. The element lies a short way above the floor and spans only half the width, and the top is open: water below it, water beside it and the surface are three different places.

1. aState State how the density of the water immediately above the element in Fig. 5.1 changes when the element is switched on, and state why it changes.

   recall2 marks

   Check answer · 2 marks

   1. the density decreases (1)
   2. the water expands when it is heated, so the same mass now occupies a larger volume (1)

2. bExplain The element drawn in Fig. 5.1 spans only half the width of the tank. Explain how the water in the half of the tank the element does not reach is warmed.

   routine3 marks

   Check answer · 3 marks

   1. water heated by the element becomes less dense and rises above it (1)
   2. it spreads sideways below the surface, while cooler, denser water sinks in the far half of the tank and flows back along the floor towards the element (1)
   3. the circulating convection current carries warmed water, and the energy it holds, into every part of the tank (1)

3. cSuggest The tank in Fig. 5.1 is open at the top. Suggest two ways in which the heated water loses energy from that open surface, and suggest one change to the tank that would reduce the loss.

   demanding3 marks

   Check answer · 3 marks

   1. evaporation: the more energetic molecules escape from the surface and carry energy away, leaving the slower ones behind (1)
   2. energy passes into the air above the surface, which is warmed and carried away by convection, or is radiated from the surface (1)
   3. fit a lid or an insulating cover over the open top (1)

4. dExplain The element is switched off after a long time. A student says that the hottest water must be at the bottom of the tank, since that is where the element is. Using Fig. 5.1, explain why the student is wrong, and state where the coolest water lies.

   top of the paper4 marks

   Check answer · 4 marks

   1. the hottest water is at the top of the tank (1)
   2. all through the heating the warmed water above the element rose and collected there, while cooler water sank (1)
   3. the coolest water is the layer trapped between the element and the floor of the tank (1)
   4. warmed water cannot sink to reach it, and water is a poor conductor, so energy reaches that layer only very slowly (1)

Reading cue. The lamp shines in from the side through the lens while the microscope looks straight down, at right angles to the beam, and the specks in view are smoke, not air molecules.

1. aState State the purpose of the converging lens drawn between the lamp and the cell in Fig. 7.1.

   recall2 marks

   Check answer · 2 marks

   1. it brings the light together, focusing the beam inside the cell (1)
   2. so that the smoke particles are lit brightly enough to be seen as separate specks (1)

2. bExplain Explain why the microscope in Fig. 7.1 is placed vertically above the cell rather than in line with the lamp.

   routine3 marks

   Check answer · 3 marks

   1. the beam from the lamp crosses the cell sideways and does not travel up into the microscope (1)
   2. only light scattered from the smoke particles travels upwards into the microscope (1)
   3. the specks are then seen bright against a dark background, whereas looking along the beam would flood the view with light from the lamp (1)

3. cExplain The cell in Fig. 7.1 is warmed gently. Explain the change seen in the movement of the specks.

   demanding3 marks

   Check answer · 3 marks

   1. the specks move faster and change direction more often, so the movement looks more violent (1)
   2. at the higher temperature the air molecules have greater average kinetic energy and move faster (1)
   3. each impact on a smoke particle is therefore harder and impacts arrive more frequently (1)

4. dSuggest A student looking down the microscope says that the specks being watched are the air molecules themselves. Suggest two pieces of evidence, one from the apparatus in Fig. 7.1 and one from what is seen, that show this cannot be so.

   top of the paper4 marks

   Check answer · 4 marks

   1. the specks can be seen through an ordinary light microscope, so they are far larger than molecules, which are much too small to be seen in this way (1)
   2. the cell held air before the smoke was introduced and nothing was visible until the smoke was added (1)
   3. the specks move in short, sudden, random jerks rather than smoothly (1)
   4. this is what is expected of a particle being struck unequally on opposite sides by very many smaller particles that cannot themselves be seen (1)

Reading cue. The vacuum gap runs down both sides and under the base but stops at the sealed neck; the stopper and the small supports are the only solid links to the outer case.

1. aIdentify Identify the two parts drawn in Fig. 9.1 that provide a solid path for thermal energy out of the inner container, and state the property of the material chosen for them.

   recall2 marks

   Check answer · 2 marks

   1. the stopper closing the neck and the small supports under the base (1)
   2. both are made of insulating material, that is poor conductors of thermal energy, and their contact area is kept small (1)

2. bExplain Explain how the stopper drawn in Fig. 9.1 slows the cooling of the drink, other than by being a poor conductor.

   routine3 marks

   Check answer · 3 marks

   1. it seals the neck, so vapour and warm air above the drink cannot escape (1)
   2. evaporation removes the fastest molecules from the surface of the drink, and this cools the drink (1)
   3. with the neck closed the escaped molecules stay trapped just above the surface and many return to the liquid, so the net loss by evaporation is small, and the warm air cannot be carried away and replaced by cooler air (1)

3. cExplain The drink does not reach the stopper: Fig. 9.1 shows a space between the surface of the drink and the neck. Explain how energy still crosses this space, and suggest how filling the flask closer to the stopper would change the rate of cooling.

   demanding3 marks

   Check answer · 3 marks

   1. the warm surface of the drink emits infrared radiation across the space (1)
   2. air in the space is warmed, circulates by convection, and some liquid evaporates into it (1)
   3. filling closer to the stopper leaves a smaller space, so less evaporation and less convection are possible and the drink cools more slowly (1)

4. dSuggest The same flask is used to keep a cold drink cold on a hot day. Suggest whether the vacuum and the silvered surfaces still do useful work, and explain your answer in terms of the direction in which energy travels.

   top of the paper4 marks

   Check answer · 4 marks

   1. both still work (1)
   2. the energy now travels inwards, from the warm surroundings towards the colder drink (1)
   3. the vacuum contains no particles, so neither conduction nor convection can carry energy across the gap in either direction (1)
   4. the silvering on the outer of the two glass walls reflects the infrared radiation arriving from the surroundings back outwards, before it can cross the gap to the drink (1)

Reading cue. Read the 48 W as a rate, not an amount: the drawing fixes the energy delivered each second, and the two holes are drilled apart so the thermometer reads the block, not the heater.

1. aState State the energy transferred to the aluminium block by the heater shown in Fig. 10.1 in one second, and state the quantity the label 48 W measures.

   recall2 marks

   Check answer · 2 marks

   1. 48 J
   2. 48 W is a power, the rate at which the heater transfers energy

2. bCalculate The specific heat capacity of aluminium is 900 J/(kg °C). Calculate the time for which the heater in Fig. 10.1 must run to raise the temperature of the block by 20 °C, assuming no thermal energy escapes.

   routine2 marks

   Check answer · 2 marks

   1. E = mcΔθ = 1.0 × 900 × 20 = 18 000 J
   2. t = E/P = 18 000 / 48 = 375 s (6.25 minutes)

3. cExplain The heater and the thermometer occupy holes at opposite sides of the block. Explain why the thermometer reading rises more slowly in the first minute after switching on than it does later in the run.

   demanding3 marks

   Check answer · 3 marks

   1. energy must be conducted through the aluminium from the heater hole across to the thermometer hole
   2. this takes time, so the thermometer lags behind the temperature of the aluminium next to the heater
   3. some of the early energy also warms the heater itself and the metal immediately around it before the block is at a uniform temperature

4. dDetermine The student switches the heater off at the end of the run and finds that the block cools by 0.60 °C in the next minute. Determine the rate at which the block was losing thermal energy at the end of the run, and hence determine the percentage of the heater's 48 W that was actually raising the temperature of the block at that moment.

   top of the paper4 marks

   Show a hint

   The block cools at the same rate whether or not the heater is on, so the cooling run measures the loss that was happening during the heating run.

   Check answer · 4 marks

   1. energy lost in 60 s = mcΔθ = 1.0 × 900 × 0.60 = 540 J
   2. rate of loss = 540/60 = 9.0 W
   3. useful power = 48 − 9.0 = 39 W
   4. percentage = 39/48 × 100 = 81%