Heating a block of aluminium — answer key

1. HQ1 · Practice

Model answer. Block B has more internal energy, because its temperature is higher. To warm a block from 20 °C to 60 °C you must supply energy, and that energy is stored in the block as internal energy. The two blocks have the same mass of the same material, so the hotter one, B, has more internal energy. Its particles vibrate faster: their average kinetic energy is higher.

A full answer includes

• I chose “Block B has more internal energy, because its temperature is higher”.
• I explained that warming a block stores energy in it: the same block at a higher temperature has more internal energy, because its particles vibrate faster on average.

Also accept

• Block B has more internal energy; same block, higher temperature, so more internal energy (its particles vibrate faster).

Watch for

• “They have the same internal energy, because they have the same mass.” — The mass tells you how many particles there are, not how fast they move. Warming block B to 60 °C took energy, and that energy is stored in it: same mass, higher temperature, more internal energy.
• “Neither block has any internal energy until it is heated.” — Every object has internal energy: its particles are always moving, even at 20 °C. Heating adds to it.

2. HQ2 · Practice

Model answer. (1) internal: the energy supplied is stored by the particles of the block. (2) vibrate: in a solid the particles cannot move from place to place, so they move back and forth about their fixed positions. (3) kinetic: the energy of moving particles; a higher temperature means a higher average kinetic energy. (4) slower: as the block cools its particles vibrate more slowly, their average kinetic energy falls and the internal energy decreases.

A full answer includes

• (1) internal
• (2) vibrate
• (3) kinetic
• (4) slower (or more slowly)

Also accept

• internal; vibrate; kinetic; slower.

Watch for

• “(2) expand” — Each particle stays the same size when the block is heated. What changes is how fast each particle vibrates.
• “(3) potential” — The energy something has because it is moving is kinetic energy. A rise in temperature is a rise in the average kinetic energy of the particles.

3. HQ3 · Practice

Model answer. ΔE = 5.85 kJ = 5850 J. m = 0.500 kg. Δθ = 50.0 °C − 20.0 °C = 30.0 °C. c = ΔE / (mΔθ) = 5850 J ÷ (0.500 kg × 30.0 °C) = 5850 ÷ 15.0 = 390 J/(kg °C).

Checked automatically

• Specific heat capacity of copper: 390 J/(kg °C) (within ±2%)
  • Common wrong answer 234 : You divided by the final temperature, 50.0 °C. Δθ is the rise: 50.0 − 20.0 = 30.0 °C.
  • Common wrong answer 195 : You left out the mass. Divide by m × Δθ: 0.500 kg × 30.0 °C = 15.0.
  • Common wrong answer 0.39 J/(kg °C): A thousand times too small: you used 5.85 instead of 5850 J, or 500 g instead of 0.500 kg. Change kJ into J and g into kg first.
  • Common wrong answer 87750 : You multiplied. c = ΔE ÷ (m × Δθ): divide the energy by the mass and by the rise.

A full answer includes

• I changed 5.85 kJ into 5850 J and found the rise: 50.0 °C − 20.0 °C = 30.0 °C.
• I calculated c = 5850 J ÷ (0.500 kg × 30.0 °C) = 390 J/(kg °C), with the unit.

Also accept

• 390 J/(kg °C).

Watch for

• “c = 5850 ÷ (0.500 × 50.0) = 234 J/(kg °C).” — 50.0 °C is the final temperature. Use the rise in temperature: 50.0 °C − 20.0 °C = 30.0 °C.
• “c = 5.85 ÷ (0.500 × 30.0) = 0.39 J/(kg °C).” — The energy must be in joules: 5.85 kJ = 5850 J.

4. HQ4 · Practice

Options as numbered on the pupil sheet: 1 — Read the final temperature and work out the rise in temperature; 2 — Weigh the block on a balance; 3 — Switch on and heat until the joulemeter shows the chosen energy, then switch off; 4 — Put the heater and the thermometer in their holes, set the joulemeter to zero and read the starting temperature.

Model answer. First weigh the block, so that you know m. Then put the heater in one hole and the thermometer in the other, set the joulemeter to zero and read the starting temperature, before any energy goes in. Switch on and heat until the joulemeter shows the chosen energy, such as 30.0 kJ, and switch off. Read the final temperature (the highest reading), subtract the starting temperature to get Δθ, change the energy into joules and calculate c = ΔE / (mΔθ).

A full answer includes

• Step 1: weigh the block on a balance.
• Step 2: put the heater and the thermometer in their holes, set the joulemeter to zero and read the starting temperature.
• Step 3: switch on and heat until the joulemeter shows the chosen energy, then switch off.
• Step 4: read the final temperature and work out the rise in temperature; then calculate c = ΔE / (mΔθ).

Also accept

• weigh the block; heater and thermometer in, joulemeter to zero, read the starting temperature; heat to the chosen energy and switch off; read the final temperature, work out the rise and c.

Watch for

• “Read the starting temperature after switching on.” — Once the heater is on, the block starts to warm, so a late starting reading is too high and the rise you work out is too small. Read it before any energy goes in.
• “Set the joulemeter to zero at the end.” — The joulemeter counts the energy from the moment it is set to zero, so zero it before you switch on; its final reading is then the energy supplied.

5. HQ5 · Practice

Model answer. m = 250 g = 0.250 kg. Δθ = 120 °C − 20 °C = 100 °C. ΔE = mcΔθ = 0.250 kg × 390 J/(kg °C) × 100 °C = 9750 J, about 9.8 kJ.

Checked automatically

• Energy needed: 9750 J (within ±2%)
  • Common wrong answer 9750000 : A thousand times too big: the mass must be in kilograms, 250 g = 0.250 kg.
  • Common wrong answer 11700 : You used the final temperature, 120 °C. Use the rise: 120 − 20 = 100 °C.
  • Common wrong answer 39000 : You left out the mass. Multiply by m = 0.250 kg as well.

A full answer includes

• I changed 250 g into 0.250 kg and found the rise: 120 °C − 20 °C = 100 °C.
• I calculated ΔE = 0.250 kg × 390 J/(kg °C) × 100 °C = 9750 J, with the unit.

Also accept

• 9750 J.

Watch for

• “ΔE = 250 × 390 × 100 = 9 750 000 J.” — Specific heat capacity is the energy per kilogram, so the mass must be in kilograms: 250 g = 0.250 kg.
• “ΔE = 0.250 × 390 × 120 = 11 700 J.” — 120 °C is the final temperature. Use the rise: 120 °C − 20 °C = 100 °C.

6. HQ6 · Practice

Model answer. (a) ΔE = 27.0 kJ = 27 000 J. Δθ = ΔE / (mc) = 27 000 J ÷ (2.00 kg × 450 J/(kg °C)) = 27 000 ÷ 900 = 30.0 °C. (b) That is the rise, so the final temperature is 20.0 °C + 30.0 °C = 50.0 °C.

Checked automatically

• (a) Rise in temperature: 30.0 °C (within ±0.5 °C)
  • Common wrong answer 0.03 : The energy must be in joules: 27.0 kJ = 27 000 J.
  • Common wrong answer 60 : You left out the mass. Divide by m × c = 2.00 kg × 450 J/(kg °C) = 900.
  • Common wrong answer 50 : 50.0 °C is the final temperature, which (b) asks for. The rise is what the energy gives: ΔE ÷ (mc).
• (b) Final temperature: 50.0 °C (within ±0.5 °C)
  • Common wrong answer 30 : That is the rise in temperature. The block started at 20.0 °C: add the rise to it.
  • Common wrong answer 80 : Check the rise: 27 000 J ÷ (2.00 kg × 450 J/(kg °C)) = 30.0 °C, not 60 °C. Then add it to 20.0 °C.

A full answer includes

• (a) ΔE = 27 000 J; Δθ = 27 000 J ÷ (2.00 kg × 450 J/(kg °C)) = 30.0 °C.
• (b) Final temperature = 20.0 °C + 30.0 °C = 50.0 °C.

Also accept

• (a) 30.0 °C; (b) 50.0 °C.

Watch for

• “(b) The final temperature is 30.0 °C.” — 30.0 °C is the rise in temperature. The block started at 20.0 °C, so add the rise: 50.0 °C.
• “(a) Δθ = 27 000 ÷ 450 = 60 °C.” — The energy is shared by 2.00 kg of iron. Divide by m × c = 2.00 × 450 = 900.

7. HQ7 · Practice

Options as numbered on the pupil sheet: 1 — the energy required per unit mass per unit temperature increase, in J/(kg °C); 2 — the energy supplied, in J; 3 — the mass, in kg; 4 — the rise in temperature, in °C; 5 — the temperature at the end, in °C; 6 — the internal energy of the block, in J.

Model answer. ΔE is the energy supplied, in joules (J), read from the joulemeter. m is the mass, in kilograms (kg), measured on a balance. Δθ is the rise in temperature, the final temperature minus the starting temperature, in °C. c is the specific heat capacity: the energy required per unit mass per unit temperature increase, in J/(kg °C).

A full answer includes

• ΔE: the energy supplied, in J.
• m: the mass, in kg.
• Δθ: the rise in temperature, in °C.
• c: the energy required per unit mass per unit temperature increase, in J/(kg °C).

Also accept

• ΔE – the energy supplied, in J; m – the mass, in kg; Δθ – the rise in temperature, in °C; c – the energy required per unit mass per unit temperature increase, in J/(kg °C).

Watch for

• “Δθ is the temperature at the end.” — Δ means “a change in”: Δθ is the rise in temperature, the final temperature minus the starting temperature.
• “ΔE is the internal energy of the block.” — ΔE is the energy supplied: the change in the internal energy, if none escapes. The block already had internal energy before you started.

8. HQ8 · Practice

Model answer. (a) At 0 kJ the block is at 20.0 °C and at 15 kJ at 35.0 °C, a rise of 15.0 °C. (b) At 30 kJ it is at 48.5 °C, so the last 15 kJ raised it by 48.5 − 35.0 = 13.5 °C. The same energy gave a smaller rise in the second half because the block was then much warmer than the air around it, so energy escaped from it faster and less of each kilojoule stayed in the aluminium. That is why the line bends.

Checked automatically

• (a) Rise during the first 15 kJ: 15.0 °C (within ±0.5 °C)
  • Common wrong answer 35 : 35.0 °C is the temperature at 15 kJ. The rise is that minus the starting temperature, 20.0 °C.
• (b) Rise during the last 15 kJ: 13.5 °C (within ±0.5 °C)
  • Common wrong answer 48.5 : 48.5 °C is the temperature at 30 kJ. Subtract the temperature at 15 kJ, 35.0 °C.
  • Common wrong answer 28.5 : 28.5 °C is the rise for the whole 30 kJ. Take only the last 15 kJ, from 15 kJ to 30 kJ.

A full answer includes

• (a) 35.0 °C − 20.0 °C = 15.0 °C.
• (b) 48.5 °C − 35.0 °C = 13.5 °C.
• I explained that the hotter the block got, the faster energy escaped to the air, so less of each kilojoule stayed in the block during the second half.

Also accept

• (a) 15.0 °C; (b) 13.5 °C; the hotter block loses energy to the air faster.

Watch for

• “Aluminium needs more energy for each degree when it is hot.” — The specific heat capacity of aluminium hardly changes over this range. In a jacket the same kind of block rises almost evenly: the difference comes from energy escaping to the air.
• “The heater supplies less energy as it goes on.” — The joulemeter shows that each 15 kJ is the same energy. The difference is in how much of it stays in the block.

9. HQ9 · Practice

Model answer. The two about energy: some escaped from the warm block to the air, and some warmed the heater and the thermometer. The calculation assumes that all 30 000 J counted by the joulemeter warmed the aluminium. Some escaped to the air and some warmed the heater and the thermometer, so the block’s temperature rose by less than it would have done. A smaller Δθ in c = ΔE / (mΔθ) gives a bigger c. The block’s mass does not change and the particles do not get bigger. Reading to the nearest 0.5 °C could put the rise out by about half a degree, which changes c by under 2 % and could make it a little low as easily as a little high: it cannot explain a value about 17 % too high. Energy that never reached the aluminium always pushes c up.

A full answer includes

• I chose the two about energy: some escaped from the warm block to the air, and some warmed the heater and the thermometer.
• I explained that not all the energy counted stayed in the block, so its rise in temperature was smaller than it would have been, and a smaller Δθ in c = ΔE / (mΔθ) gives a bigger c.

Also accept

• energy escaped to the air, and energy warmed the heater and the thermometer; less energy stayed in the block, so Δθ was too small and c too big.

Watch for

• “The thermometer could only be read to the nearest 0.5 °C.” — Half a degree out of 28.5 °C changes c by under 2 %, and it could make c too low as easily as too high. The measured value is about 17 % too high: something pushed it up.
• “The block lost mass as it was heated.” — Heating does not change the mass of a metal block: no particles leave it. Its mass is the same before and after.

10. HQ10 · Practice

Model answer. The particles do not get bigger: each particle of the metal keeps its size, and there are just as many of them, still vibrating about fixed places in the same regular pattern. What changes is how they move. As the block is heated they vibrate faster, so the average kinetic energy of all the particles increases, and a rise in temperature is exactly that. The energy supplied is stored by the particles, so the internal energy of the block increases. (The block does expand very slightly, because the particles end up a tiny bit further apart, but that is not what makes its temperature rise.)

A full answer includes

• I said the particles stay the same size (and there are just as many, still vibrating about fixed places in the same pattern).
• I said they vibrate faster, so the average kinetic energy of all the particles increases: that is what a rise in temperature means.
• I said the internal energy of the block increases.

Also accept

• the particles stay the same size; they vibrate faster, so their average kinetic energy rises; the internal energy increases.

Watch for

• “The particles move from place to place faster.” — In a solid the particles cannot move from place to place: each one vibrates about a fixed place. They vibrate faster as the block warms.
• “Heating adds more particles to the block.” — The mass of the block does not change when it is heated, so no particles are added. The same particles move faster.

11. HT1 · Apply

Model answer. (a) ΔE = 450 kJ = 450 000 J. Δθ = ΔE / (mc) = 450 000 J ÷ (5.0 kg × 450 J/(kg °C)) = 450 000 ÷ 2250 = 200 °C. (b) The temperature of the disc rises, so its internal energy increases: the energy of the moving car is now stored by the particles of the steel, which vibrate faster, so their average kinetic energy is higher. (c) The 10 kg disc needs twice as much energy for each degree, so the same 450 000 J raises its temperature by half as much: 100 °C. A disc that stays cooler is less likely to overheat. (d) Not all the energy counted by the joulemeter stayed in the steel: some escaped to the air and some warmed the heater and the thermometer. The rise in temperature was therefore smaller than it should have been, and a smaller Δθ gives a larger c. Wrapping the sample in insulation, such as a polystyrene jacket, cuts the energy that escapes.

A full answer includes

• (a) Δθ = 450 000 J ÷ (5.0 kg × 450 J/(kg °C)) = 200 °C.
• (b) The temperature rises, so the internal energy increases: the particles of steel vibrate faster, with a higher average kinetic energy.
• (c) 100 °C: twice the mass needs twice the energy for each degree, so the same energy gives half the rise; the disc stays cooler.
• (d) Some energy escaped to the air or warmed the heater and the thermometer, so Δθ was too small and c too big; insulate the sample (a jacket), or put a drop of oil in the thermometer’s hole.

Also accept

• (a) 200 °C; (b) internal energy increases, particles vibrate faster, higher average kinetic energy; (c) 100 °C, stays cooler; (d) energy lost to the air and the apparatus, so Δθ too small; insulate the sample.

Watch for

• “(a) Δθ = 450 ÷ (5.0 × 450) = 0.2 °C.” — The energy must be in joules: 450 kJ = 450 000 J.
• “(c) A heavier disc warms more, because there is more steel to heat.” — More steel shares the same energy, so each kilogram gets less. Twice the mass gives half the rise in temperature.