IGCSE Physics 2.2 Thermal Properties and Temperature MCQs – 30 Practice Questions with Answers (Cambridge 0625)
Practise Section 2.2 of the Cambridge IGCSE Physics (0625) syllabus with 30 original multiple-choice questions on thermometers and temperature scales, specific heat capacity, and latent heat. Every question comes with a fully worked answer – tap Show Answer to check your reasoning before you move on.
Formula recap: energy for a temperature change: Q = mcΔT. Energy for a change of state: Q = mL. Temperature conversion: K = °C + 273. For every equation used in this section, see the full IGCSE Physics Formula Sheet.
Thermometers & Temperature Scales
Q1. What does the sensitivity of a liquid-in-glass thermometer describe?
- (A) The highest temperature it can measure
- (B) The distance the liquid thread moves for a temperature change of 1 °C
- (C) The time it takes to give a steady reading
- (D) How evenly spaced its degree marks are
Show Answer
Answer: B – A sensitive thermometer shows a large movement of the thread for a small temperature change. The highest temperature (A) is the range, (C) is the response time and (D) is the linearity – three separate properties.
Q2. Which change would make a liquid-in-glass thermometer more sensitive?
- (A) Using a wider capillary tube
- (B) Using a larger bulb and a narrower capillary tube
- (C) Making the glass wall of the bulb thicker
- (D) Using a shorter stem
Show Answer
Answer: B – A larger bulb holds more liquid, so a greater volume of liquid is pushed into the tube for each degree of rise; a narrower capillary means that volume travels further along the scale, so the thread moves more per degree.
Q3. A manufacturer wants to increase the range of a liquid-in-glass thermometer. Which change should be made?
- (A) Use a smaller bulb containing less liquid
- (B) Use a narrower capillary tube
- (C) Make the bulb wall thinner
- (D) Use a larger bulb containing more liquid
Show Answer
Answer: A – With less liquid in the bulb, the total expansion over a wide temperature range is smaller, so the thread stays on the scale at the top end. A larger bulb (D) increases sensitivity, not range – the opposite design choice from Q2.
Q4. A thermocouple is chosen to measure the temperature inside an industrial furnace at about 1200 °C. The main reason is that a thermocouple
- (A) needs no temperature scale
- (B) is filled with mercury
- (C) has a junction made of glass
- (D) has a very wide range and responds very quickly to temperature changes
Show Answer
Answer: D – A thermocouple (a junction of two different metals) works over a very wide range, well above 1000 °C where a liquid-in-glass thermometer would fail, and its small junction reaches the surrounding temperature almost instantly.
Q5. When the Celsius scale is set up, the upper fixed point (the steam point) is given the value
- (A) 0 °C
- (B) 100 °C
- (C) 212 °C
- (D) 373 °C
Show Answer
Answer: B – The lower fixed point (the ice point, pure melting ice) is 0 °C and the upper fixed point (the steam point, steam from pure water boiling at standard atmospheric pressure) is 100 °C. The interval between them is divided into 100 equal degrees.
Q6. Normal body temperature is 37 °C. What is this temperature on the Kelvin scale?
- (A) 236 K
- (B) 273 K
- (C) 300 K
- (D) 310 K
Show Answer
Answer: D – K = °C + 273, so 37 + 273 = 310 K.
Q7. The temperature of a room rises by 25 °C during the day. This temperature rise is equal to
- (A) 25 K
- (B) 248 K
- (C) 278 K
- (D) 298 K
Show Answer
Answer: A – A temperature difference of 1 °C is exactly the same size as a difference of 1 K, so a rise of 25 °C is a rise of 25 K. Only the zero points of the two scales differ; you add 273 only when converting a temperature itself, not a difference.
Q8. At an Antarctic research station the air temperature falls to −55 °C. Which liquid is most suitable for the liquid-in-glass thermometer used outside?
- (A) Mercury, which freezes at −39 °C
- (B) Water, which freezes at 0 °C
- (C) Coloured alcohol, which freezes below −100 °C
- (D) Any of these liquids would work equally well
Show Answer
Answer: C – Mercury and water would both be solid at −55 °C, so their thermometers could not work. Alcohol stays liquid down to below −100 °C, which is why alcohol thermometers are used in very cold places. (Mercury is preferred for higher temperatures because it does not wet the glass, has a higher boiling point and expands more uniformly.)
Q9. A thermocouple connected to a voltmeter is used to monitor a furnace from a control room some distance away. This remote reading is possible because
- (A) the junction must be read by eye, like a liquid thread
- (B) the two wires act as electrical insulators
- (C) the junction produces a voltage that can be measured by a meter placed far from the hot junction
- (D) the thermocouple contains a liquid that expands up a tube
Show Answer
Answer: C – The hot junction generates a small voltage that depends on its temperature. The meter measuring that voltage can sit at the other end of long wires, well away from the heat – something a liquid-in-glass thermometer can never do.
Q10. The glass wall of the bulb of a liquid-in-glass thermometer is made very thin so that
- (A) the thermometer has a larger range
- (B) the liquid inside expands by a greater amount
- (C) the scale becomes linear
- (D) heat is conducted quickly through the glass to the liquid, so the thermometer responds quickly
Show Answer
Answer: D – Thin glass lets heat pass rapidly to the liquid inside, so the reading settles quickly when the thermometer is placed somewhere hotter or colder. It does not change how much the liquid expands.
Specific Heat Capacity: Q = mcΔT
Q11. The specific heat capacity of a substance is the energy needed to
- (A) raise the temperature of 1 kg of the substance by 1 °C
- (B) melt 1 kg of the substance
- (C) raise the temperature of any mass of the substance by 1 °C
- (D) convert 1 kg of the substance into a gas
Show Answer
Answer: A – By definition, the specific heat capacity c is the energy per kilogram per degree, which is why the equation is Q = mcΔT. Option (B) describes the specific latent heat of fusion and (D) the specific latent heat of vaporisation.
Q12. What is the unit of specific heat capacity?
- (A) J/kg
- (B) J/°C
- (C) J/(kg °C)
- (D) J kg/°C
Show Answer
Answer: C – Rearranging Q = mcΔT gives c = Q/(mΔT), so the unit is joules divided by (kilograms × degrees Celsius). J/kg (A) is the unit of specific latent heat.
Q13. Calculate the energy needed to raise the temperature of 2.0 kg of water from 20 °C to 40 °C. (Specific heat capacity of water = 4200 J/(kg °C))
- (A) 84 000 J
- (B) 168 000 J
- (C) 420 000 J
- (D) 1 680 000 J
Show Answer
Answer: B – ΔT = 40 − 20 = 20 °C. Q = mcΔT = 2.0 × 4200 × 20 = 168 000 J.
Q14. A 0.50 kg aluminium block (c = 900 J/(kg °C)) is supplied with 9000 J of energy and no energy is lost. What is its temperature rise?
- (A) 20 °C
- (B) 2.0 °C
- (C) 200 °C
- (D) 10 °C
Show Answer
Answer: A – ΔT = Q/(mc) = 9000/(0.50 × 900) = 9000/450 = 20 °C.
Q15. A 1.5 kg block of metal is heated with 9450 J of energy and its temperature rises by 14 °C. What is the specific heat capacity of the metal?
- (A) 675 J/(kg °C)
- (B) 450 J/(kg °C)
- (C) 6300 J/(kg °C)
- (D) 45 J/(kg °C)
Show Answer
Answer: B – c = Q/(mΔT) = 9450/(1.5 × 14) = 9450/21 = 450 J/(kg °C). The trap answers come from forgetting one factor: 675 ignores the mass, and 6300 ignores the temperature rise.
Q16. What mass of copper (c = 390 J/(kg °C)) can be heated through 40 °C by 7800 J of energy?
- (A) 0.05 kg
- (B) 0.50 kg
- (C) 2.0 kg
- (D) 5.0 kg
Show Answer
Answer: B – m = Q/(cΔT) = 7800/(390 × 40) = 7800/15 600 = 0.50 kg.
Q17. Water is used as the coolant in a car engine mainly because
- (A) water has a very high specific heat capacity, so it absorbs a large amount of energy for only a small rise in its own temperature
- (B) water is a very good electrical conductor
- (C) water does not expand when it is heated
- (D) water has a very low density
Show Answer
Answer: A – With c = 4200 J/(kg °C), every kilogram of water flowing through the engine carries away a large amount of energy while its own temperature rises by only a few degrees – exactly what a coolant needs to do.
Q18. Equal masses of water and oil are heated by identical immersion heaters for the same time, so both liquids receive the same amount of energy. The oil reaches a higher final temperature than the water. This shows that, compared with water, the oil has
- (A) a higher boiling point
- (B) a greater density
- (C) a lower specific heat capacity
- (D) a higher specific heat capacity
Show Answer
Answer: C – With the same mass and the same energy input, ΔT = Q/(mc): the liquid with the larger temperature rise must have the smaller c. Water’s unusually high specific heat capacity makes it heat up slowly.
Q19. Coastal towns usually have smaller temperature changes between day and night than places far inland. This is mainly because
- (A) the sea is deeper than the land is high
- (B) sand has a higher specific heat capacity than water
- (C) water has a high specific heat capacity, so the sea warms up and cools down much more slowly than the land
- (D) sea water reflects all the sunlight that falls on it
Show Answer
Answer: C – Soil and rock have much lower specific heat capacities than water, so the land heats up quickly by day and cools quickly at night. The sea’s temperature stays far steadier, and it moderates the temperature of the air above the nearby coast.
Q20. A student determines the specific heat capacity of a metal block using an electrical heater fitted into a hole in the block. Which set of measurements must she make?
- (A) The mass of the block, the current in the heater, the voltage across it, the heating time and the temperature change of the block
- (B) Only the mass of the block and its final temperature
- (C) The volume of the block and its density
- (D) Only the current in the heater and its resistance
Show Answer
Answer: A – The electrical energy supplied is E = IVt (current × voltage × time), and then c = IVt/(mΔT). The block is also wrapped in insulation (lagged) to reduce energy losses to the surroundings, which would otherwise make the calculated value of c too large.
Latent Heat & Heating Curves [mostly Supplement]
Q21. The specific latent heat of fusion of a substance is the energy needed to [Supplement]
- (A) change 1 kg of the substance from solid to liquid without a change in temperature
- (B) raise the temperature of 1 kg of the substance by 1 °C
- (C) freeze any mass of the substance
- (D) melt 1 g of the substance at any temperature
Show Answer
Answer: A – “Specific” means per kilogram, and a change of state takes place at a single temperature (the melting point) with no temperature change. Option (B) is the definition of specific heat capacity.
Q22. The specific latent heat of vaporisation of water is the energy needed to [Supplement]
- (A) boil any mass of water completely away
- (B) raise the temperature of 1 kg of water by 1 °C
- (C) convert 1 kg of water into steam without a change in temperature
- (D) convert 1 kg of steam into water by heating it
Show Answer
Answer: C – It is the energy per kilogram for the change liquid → gas at the boiling point, at constant temperature, used in Q = mL. Option (D) describes the reverse change (condensing), which releases the same amount of energy per kilogram.
Q23. How much energy is needed to melt 0.50 kg of ice at 0 °C? (Specific latent heat of fusion of ice = 3.4 × 105 J/kg) [Supplement]
- (A) 68 000 J
- (B) 340 000 J
- (C) 680 000 J
- (D) 170 000 J
Show Answer
Answer: D – Q = mL = 0.50 × 3.4 × 105 = 1.7 × 105 J = 170 000 J. No temperature change is involved, so Q = mcΔT does not apply.
Q24. How much energy is needed to boil away 0.20 kg of water at 100 °C? (Specific latent heat of vaporisation of water = 2.3 × 106 J/kg) [Supplement]
- (A) 4.6 × 104 J
- (B) 1.15 × 106 J
- (C) 2.3 × 106 J
- (D) 4.6 × 105 J
Show Answer
Answer: D – Q = mL = 0.20 × 2.3 × 106 = 4.6 × 105 J. Compare Q23: vaporisation needs far more energy per kilogram than fusion, because the molecules must be separated completely, not just loosened.
Q25. A heater supplies 6.8 × 105 J to crushed ice at 0 °C. What mass of ice melts? (Specific latent heat of fusion of ice = 3.4 × 105 J/kg) [Supplement]
- (A) 0.50 kg
- (B) 1.0 kg
- (C) 2.0 kg
- (D) 0.20 kg
Show Answer
Answer: C – m = Q/L = (6.8 × 105)/(3.4 × 105) = 2.0 kg.
Q26. A beaker of crushed ice is heated steadily. While the ice is melting, the thermometer in the beaker stays at 0 °C even though energy is still being supplied. During melting, the energy is used to
- (A) break the bonds holding the water molecules in the solid structure, increasing the potential energy of the molecules
- (B) raise the average kinetic energy of the molecules
- (C) warm the thermometer only
- (D) escape as sound
Show Answer
Answer: A – Temperature measures the average kinetic energy of the molecules, and that stays constant during a change of state. The supplied energy instead separates the molecules against the intermolecular forces, increasing their potential energy – which is why the temperature only starts rising again once all the ice has melted.
Q27. A solid is heated uniformly and a temperature-time graph is plotted. The graph rises, then becomes horizontal for a while, then rises again. During the horizontal part, the substance is
- (A) cooling down
- (B) changing state at a constant temperature
- (C) no longer absorbing any energy
- (D) at room temperature
Show Answer
Answer: B – The flat (plateau) section is a change of state, here melting: energy is still being absorbed, but all of it goes into the latent heat while the temperature stays fixed. The same plateaus appear on a cooling curve, when the change of state releases energy instead.
Q28. Steam at 100 °C produces a more serious burn than the same mass of boiling water at 100 °C landing on the skin because [Supplement]
- (A) steam is at a higher temperature than boiling water
- (B) steam contains stored chemical energy
- (C) steam is invisible, so the burn is not noticed
- (D) the steam first condenses on the skin, releasing its latent heat of vaporisation, and the hot water formed then also cools on the skin
Show Answer
Answer: D – Both are at 100 °C, but condensing steam releases 2.3 × 106 J/kg of latent heat directly onto the skin, on top of the energy the resulting hot water gives out as it cools. That is far more energy than the same mass of boiling water alone can deliver.
Q29. Evaporation of sweat cools the skin. This cooling happens because
- (A) the most energetic molecules escape from the surface of the liquid, so the average kinetic energy of the molecules left behind falls
- (B) every molecule in the liquid loses the same amount of energy
- (C) cold air is drawn into the liquid
- (D) the escaping molecules give their energy back to the liquid
Show Answer
Answer: A – Evaporation takes place at the surface at any temperature: the fastest-moving molecules have enough energy to break free, so the average kinetic energy of those remaining decreases. Since temperature is a measure of that average kinetic energy, the liquid (and the skin beneath it) cools.
Q30. A liquid boils at a constant temperature while energy is supplied continuously. The internal energy of the substance increases during boiling because [Supplement]
- (A) the average kinetic energy of the molecules increases
- (B) neither the kinetic energy nor the potential energy of the molecules changes
- (C) the molecules are moved further apart against the intermolecular forces, so their potential energy increases
- (D) the mass of the liquid increases
Show Answer
Answer: C – Temperature is a measure of the average kinetic energy of the molecules, and it stays constant during boiling, so (A) is impossible. The extra internal energy is stored as potential energy as the molecules are pulled apart against the intermolecular forces to form a gas.
Keep practising Physics: browse all boards and topics on the Physics MCQs page, and keep the IGCSE Physics Formula Sheet open while you revise – it collects every equation for Cambridge 0625 in one place.
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