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JEE Main 2014 · 11 Apr · Shift 1, Physics question 11

A hot body, obeying Newton's law of cooling is cooling down from its peak value 80°C to an ambient temperature of 30°C. It takes 5 minutes in cooling down from 80°C to 40°C. How much time will it take to cool down from 62°C to 32°C ?

(Given ln⁡2=0.693\ln2=0.693, ln⁡5=1.609\ln5=1.609)

Choose one option
Answer and worked solution

Official answer: B

Newton's law: θ−30=(θ0−30)e−kt\theta-30=(\theta_0-30)e^{-kt}.

  • 80→4080\to40 in 5 min: ln⁡5010=5k\ln\frac{50}{10}=5k, so k=ln⁡55k=\frac{\ln5}5.
  • 62→3262\to32: ln⁡322=ln⁡16=4ln⁡2\ln\frac{32}{2}=\ln16=4\ln2.

t=4ln⁡2k=5×2.7721.609≈8.6 min.t=\frac{4\ln2}{k}=\frac{5\times2.772}{1.609}\approx8.6\text{ min}.

Topics:Newton's law of cooling
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