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S4 Physics heat transfer paper 2: ratios, convection and radiation
S4 Physics heat transfer paper 2: ratios, convection and radiation
Ten items apply the conduction formula to ratios, reverse work and double glazing, and demand density reasoning.
Direct answers Items 01 to 03 cover ratios and conductivity, 04 to 06 area and convection, 07 to 09 double glazing and surfaces. Allow 25 minutes.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Physics Heat Transfer |
| Level | S4 |
02 Questions (10 items)
- FoundationThe thickness of a board rises from 4 mm to 12 mm, all else unchanged. What fraction of the original rate remains?
- FoundationA copper plate with k = 400 W/(m K), area 2.0 m2, 50 K difference and thickness 10 mm. Find the rate.
- FoundationA board of area 0.5 m2 and thickness 0.05 m carries 500 W across 25 K. Find its thermal conductivity.
- IntermediateA board with k = 0.6 W/(m K) and thickness 0.03 m must carry 900 W across 30 K. Find the area needed.
- IntermediateWhy are air conditioners fitted high on the wall? Answer with density and circulation.
- IntermediateWhy are heaters placed near the floor?
- ChallengeDouble glazing has 5 mm glass (k = 1.0) and 10 mm air (k = 0.026), area 1 m2, 20 K difference. Find the rate, and that of 5 mm glass alone.
- ChallengeWhich absorbs more radiation, a rough black surface or a smooth silver one, and why?
- ChallengeWhich two mechanisms does the vacuum layer of a flask block, and which layer deals with radiation?
- Error redoWith k = 1.0 W/(m K), area 500 cm2, 20 K difference and thickness 0.02 m, a student answers 500000 W. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | One third | The rate is inversely proportional to thickness, and 12 / 4 = 3, so one third remains. |
| 2 | 4000000 W | 10 mm = 0.01 m, so Q/t = 400 x 2 x 50 / 0.01 = 4000000 W. |
| 3 | 2.0 W/(m K) | k = (Q/t) x L / (A delta T) = 500 x 0.05 / 12.5 = 2.0 W/(m K). |
| 4 | 1.5 m2 | A = (Q/t) x L / (k delta T) = 900 x 0.03 / 18 = 1.5 m2. |
| 5 | Cool air is denser and sinks, so a high unit drives circulation through the whole room | Cool air is denser and sinks, setting up circulation with the warmer air below, so a high unit covers the room. |
| 6 | Warm air is less dense and rises, so a low heater circulates the whole room | Heated air expands, becomes less dense and rises, so a low heater drives the circulation. |
| 7 | about 51.3 W; 4000 W for single glazing | Thermal resistances add: 0.005 / 1.0 + 0.01 / 0.026 = 0.3896, so Q/t = 20 / 0.3896 is about 51.3 W. |
| 8 | The rough black one, which also emits more | A rough black surface has high absorptivity and is also a good emitter, which is why flask walls are silvered. |
| 9 | It blocks conduction and convection; the silvered wall reflects radiation | A vacuum has no particles, so conduction and convection cannot occur; the silvered surface reflects radiation back. |
| 10 | 50 W | 500 cm2 = 0.05 m2, so Q/t = 1.0 x 0.05 x 20 / 0.02 = 50 W; the slip is treating cm2 as m2. |
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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