A body takes 4 min. to cool from $61^{\circ}\mathrm{C}$ to $59^{\circ}\mathrm{C}$. If the temperature of the surroundings is $30^{\circ}\mathrm{C}$, the time taken by the body to cool from $51^{\circ}\mathrm{C}$ to $49^{\circ}\mathrm{C}$ is:

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Heat & Thermodynamics

A body takes 4 min. to cool from

6 1^{\circ} C

5 9^{\circ} C

. If the temperature of the surroundings is

3 0^{\circ} C

, the time taken by the body to cool from

5 1^{\circ} C

4 9^{\circ} C

is:

(A)6 min.✓ Correct
(B)3 min.
(C)4 min.
(D)8 min.

Answer: (A)

Solution

Over a small drop Newton's law of cooling may be written with the mean excess temperature,

\frac{Δ θ}{Δ t} = k (θ_{a v} - θ_{0})

. For the fall from

6 1^{\circ}

C to

5 9^{\circ}

C the mean is

6 0^{\circ}

C, so

\frac{2}{4} = k (60 - 30) \Rightarrow k = \frac{1}{60} mi n^{- 1}

For the fall from

5 1^{\circ}

C to

4 9^{\circ}

C the mean is

5 0^{\circ}

C, so

\frac{2}{t} = \frac{1}{60} (50 - 30) \Rightarrow t = 6

The body takes

6

minutes for the second fall.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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