JEE Main 2016 · Physics · Calorimetry and Heat Transfer · easy

A body cools from 80^circC to 60^circC in 10 minutes in a room at 20^circC. Assuming Newton's law of cooling holds exactly, the time it takes to cool…

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A body cools from

80^\circ\mathrm{C}

60^\circ\mathrm{C}

10

minutes in a room at

20^\circ\mathrm{C}

. Assuming Newton's law of cooling holds exactly, the time it takes to cool further from

60^\circ\mathrm{C}

40^\circ\mathrm{C}

is nearly

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Answer: (C) $17.1\ \mathrm{min}$

Under Newton's law the excess temperature decays exponentially:

\theta=\theta_0e^{-kt}

. The excess falls from

60^\circ

40^\circ

10

min, so

e^{10k}=\dfrac{60}{40}=1.5

and

k=\dfrac{\ln1.5}{10}

. The next stage is excess

40^\circ\to20^\circ

, a factor of

2

, so

t=\dfrac{\ln2}{k}=\dfrac{10\ln2}{\ln1.5}=\dfrac{10(0.693)}{0.405}=17.1\ \mathrm{min}

. The linear (average temperature) approximation gives about

16.7

min, which is why the exponential form must be used when the question says 'exactly'.

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake