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Class 11 Physics Chapter 10: Thermal Properties of Matter case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
Meera pours tea into a steel tumbler in a room kept at 25 °C and reads a thermometer in the tea every two minutes, stirring gently. Her readings are in the table. She wants to know whether the cooling follows Newton's law, and plans to plot her data in a form that would make the law easy to test.
Time (min)0246
Temperature of tea (°C)85.073.063.455.7
Part 1 · 1 mark
From the table, the tea cools fastest
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Answer: (a) in the first 2 minutes

The falls are 12.0, 9.6 and 7.7 °C in successive 2-minute intervals; the rate is greatest at first, when the excess over the room is largest.

Part 2 · 1 mark
Meera's data would give a straight line with a negative slope if she plotted
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Answer: (b) ln(T − 25 °C) against t

Integrating Newton's law gives ln(T₂ − T₁) = −Kt + c, so ln of the excess temperature against time is a straight line of slope −K.

Part 3 · 2 marks
Show from the data that the excess temperature over the room falls by the same factor in each 2-minute interval, and use this to predict the temperature of the tea at 8 minutes.
Answer and marking scheme

Excesses: 60.0, 48.0, 38.4, 30.7 °C. Ratios: 48.0/60.0 = 0.80, 38.4/48.0 = 0.80, 30.7/38.4 = 0.80; the same factor, as T₂ − T₁ = C′e^(−Kt) requires. At 8 min the excess is 30.7 × 0.80 ≈ 24.6 °C, so the tea is at about 25 + 24.6 = 49.6 °C.

  • excesses and constant ratio 0.80 (1)
  • about 49.6 °C at 8 min (1)
Or
Using the average temperature of the tea in the first 2 minutes, estimate the constant K in the rate of cooling = K × (excess temperature), and use it to estimate how fast the tea cools when it is at 45 °C.
Answer and marking scheme

In the first 2 minutes the tea falls 12.0 °C, a rate of 6.0 °C per minute, at an average of 79 °C, i.e. an excess of 54 °C. K = 6.0/54 ≈ 0.111 per minute. At 45 °C the excess is 20 °C, so the rate is about 0.111 × 20 ≈ 2.2 °C per minute.

  • K ≈ 0.111 per minute from 6.0 °C/min at 54 °C excess (1)
  • about 2.2 °C per minute at 45 °C (1)
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