JEE Main 2019 · Physics · Calorimetry and Heat Transfer · medium

A body of water cooling in a room at 20^circC follows Newton's law of cooling. Its temperature falls from 80^circC to 60^circC in 10 minutes. The…

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A body of water cooling in a room at

20^\circ\mathrm{C}

follows Newton's law of cooling. Its temperature falls from

80^\circ\mathrm{C}

60^\circ\mathrm{C}

10

minutes. The time it will then take to cool from

60^\circ\mathrm{C}

50^\circ\mathrm{C}

is nearly

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Answer: (A) $7.1$ minutes

Newton's law gives

\dfrac{dT}{dt}=-k(T-T_s)

, so the excess temperature decays exponentially:

\ln\dfrac{T_1-T_s}{T_2-T_s}=kt

. For the first stage,

\ln\dfrac{60}{40}=10k

, so

k=0.0405\ \mathrm{min^{-1}}

. For the second stage,

t=\dfrac1k\ln\dfrac{40}{30}=\dfrac{0.2877}{0.0405}\approx7.1\ \mathrm{min}

. A linear estimate using the first average rate of

2^\circ\mathrm{C\,min^{-1}}

would give

5

minutes, which ignores that the cooling rate falls as the water approaches room temperature.

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