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Question 30874
JEE Physics · Hard

Elasticity, Thermal Expansion, Calorimetry and Heat Transfer: Newton's Cooling MCQ

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Newton's CoolingHardQuestion 30874

Question

A body cools from 80°C to 60°C in 5 min in a room at 20°C. According to Newton's law of cooling, the time to cool from 60°C to 40°C is:
  1. A
    8.55 min
    Correct
  2. B
    5 min
  3. C
    7.5 min
  4. D
    10 min

Correct answer

8.55 min

Explanation

The excess temperature decays exponentially. From the first interval, e−5k=(60−20)/(80−20)=2/3e^{-5k}=(60-20)/(80-20)=2/3. For the second, e−kt=(40−20)/(60−20)=1/2e^{-kt}=(40-20)/(60-20)=1/2. Thus t=5ln⁡2/ln⁡(3/2)≈<strong>8.55 min</strong>t=5\ln2/\ln(3/2)≈<strong>8.55\text{ min}</strong>.

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