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S4 Physics: Heat and Gases Practice Paper 4
S4 Physics: Heat and Gases Practice Paper 4
Paper four opens with a quarter kilo, a 15 to 40 C rise and Boyle law, moves to heating time, a 400 to 600 K Charles step and mixing three masses, then reverses for specific heat capacity and 75% efficiency, ending with the gram slip.
Quick answer 21 000 J, 210 000 J, 2 L, 16 700 J, 16.8 s, 4.5 m3, 80 C, 1 400 J/(kg K), 89.6 s, 25 200 J.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Physics: Heat and Gases – Tutor Lesson Plan |
| Level | S4 |
02 Questions (10 items)
- Foundation0.25 kg of water rises by 20 K. Find the energy absorbed. (c = 4 200 J/(kg K))
- Foundation2 kg of water is heated from 15 C to 40 C. Find the energy absorbed.
- Foundation4 L of gas at 1 atm is compressed to 2 atm at constant temperature. Find the new volume.
- Foundation0.05 kg of ice melts completely. Find the energy absorbed. (latent heat of fusion 3.34 x 10^5 J/kg)
- IntermediateA 2 000 W heater raises 0.4 kg of water by 20 K. Find the time needed.
- IntermediateA gas at 400 K and 3 m3 is heated to 600 K at constant pressure. Find the new volume.
- Intermediate0.4 kg of water at 90 C is mixed with 0.1 kg at 40 C. Find the final temperature.
- ChallengeA metal of mass 0.5 kg absorbs 42 000 J and rises 60 K. Find its specific heat capacity.
- ChallengeA 1 500 W heater of 75% efficiency raises 0.8 kg of water by 30 K. Find the time needed.
- Error redo200 g of water rises by 30 K. Find the energy absorbed. A common mistake gives 25 200 000 J; why is that wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 21 000 J (0.25 x 4 200 x 20) | E = m c dT, a straight product of the three values. |
| 2 | 210 000 J (2 x 4 200 x 25) | Get the 25 K rise first, then multiply by mass and specific heat capacity. |
| 3 | 2 L (p1 V1 = p2 V2) | At constant temperature, doubling the pressure halves the volume. |
| 4 | 16 700 J (0.05 x 3.34 x 10^5) | Melting happens at a fixed temperature, so use latent heat rather than c. |
| 5 | 33 600 J, so 16.8 s | Get the energy first, then divide by the power. |
| 6 | 4.5 m3 (V is proportional to T, a factor of 1.5) | Use kelvin and scale the volume by the same factor. |
| 7 | 80 C ((0.4 x 90 + 0.1 x 40) / 0.5) | Divide the total energy by the total mass; the capacity cancels. |
| 8 | 1 400 J/(kg K) (42 000 / (0.5 x 60)) | Rearrange E = m c dT so mass and rise form the divisor. |
| 9 | 100 800 J useful, 134 400 J supplied, so 89.6 s | Divide the useful energy by the efficiency to get what the heater really supplies. |
| 10 | 25 200 J (0.2 x 4 200 x 30); treating 200 g as 200 kg inflates it 1 000 times | Convert first: 200 g is 0.2 kg, and units must match before substitution. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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