2020 A/L Physics Past Paper · Question 34 (English Medium)

Newton's law of cooling — time to cool from 44 °C to 36 °C

A hot body, suspended in a room at 30∘C30^\circ\text{C}, cools from 60∘C60^\circ\text{C} to 50∘C50^\circ\text{C} in 5.05.0 minutes. How long will it take the same body, under the same conditions, to cool further from 44∘C44^\circ\text{C} to 36∘C36^\circ\text{C}?

  1. (1) 1010 minutes
  2. (2) 12.512.5 minutes
  3. (3) 1515 minutes
  4. (4) 2020 minutes
  5. (5) 2525 minutes
Answer and explanation

Answer: (1) 1010 minutes

Given information

Given: room θR=30∘C\theta_R = 30^\circ\text{C}; 60→50∘C60 \to 50^\circ\text{C} takes 5.05.0 minutes; under the same conditions, 44→36∘C44 \to 36^\circ\text{C}.

What the question asks

We must find: the time taken to cool from 44∘C44^\circ\text{C} to 36∘C36^\circ\text{C}.

Relevant theory

The rate of loss of heat of a body, when it is cooling under conditions of forced convection, is directly propertional to the difference in temperature between the body and its surroundings.

Required equations

dθdt=K(θ−θR)\frac{d\theta}{dt} = K(\theta - \theta_R)

ΔθΔt≈K(θˉ−θR)\frac{\Delta\theta}{\Delta t} \approx K(\bar\theta - \theta_R)

Applying it to this question

Plan: (1) find KK from the first cooling (mean excess temperature 25∘C25^\circ\text{C}).

(2) find the rate at the second cooling's mean excess temperature 10∘C10^\circ\text{C}.

(3) time=drop÷rate. Since the body and surroundings are unchanged(3)\,time = drop \div rate.\,Since\,the\,body\,and\,surroundings\,are\,unchanged K=kAmcK = \dfrac{kA}{mc} stays constant.

Reasoning and calculation

2.02.0

60−505.0\frac{60 - 50}{5.0}

2525

60+502−30\frac{60 + 50}{2} - 30

0.0800.080

K=2.025K = \frac{2.0}{25}

1010

44+362−30\frac{44 + 36}{2} - 30

0.800.80

0.080×100.080 \times 10

t=44−360.80t = \frac{44 - 36}{0.80}

Checks

Unit check: KK has units ∘C/min∘C=min−1\dfrac{^\circ\text{C/min}}{^\circ\text{C}} = \text{min}^{-1}; ∘C∘C/min=min\dfrac{^\circ\text{C}}{^\circ\text{C/min}} = \text{min} ✓.

Physical plausibility: the second drop is smaller (88 vs 10∘C10^\circ\text{C}), but the rate is 2.52.5 times slower — so the time is longer: 5×2.5×0.8=105 \times 2.5 \times 0.8 = 10 minutes ✓.

(Solving the full exponential gives ≈10.4\approx 10.4 minutes — option (1) is still nearest.)

8/(10/t1∗10/25)8 / (10 / t1 * 10 / 25)

Conclusion

t=80.8=10 mint = \frac{8}{0.8} = 10\ \text{min}

The rate is proportional to the excess temperature: 2.0→0.8 ∘C2.0 \to 0.8\ ^\circ\text{C}/min; an 8∘C8^\circ\text{C} drop takes 1010 minutes — the answer is (1).

Assuming the rate is proportional to the body's own temperature.

Newton's law: the rate is proportional to (θ−θR)(\theta - \theta_R) — the excess over the surroundings.

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