A body cools from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 5 minutes. The temperature of the surrounding is $20^{\circ} \mathrm{C}$. The time it takes to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is:

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Heat & Thermodynamics

A body cools from

8 0^{\circ} C

6 0^{\circ} C

in 5 minutes. The temperature of the surrounding is

2 0^{\circ} C

. The time it takes to cool from

6 0^{\circ} C

4 0^{\circ} C

is:

(A)500 s✓ Correct
(B) $\frac{25}{3} s$
(C)450 s
(D)420 s

Answer: (A)

Solution

By Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the body and surroundings.

For the first interval (80°C to 60°C in 5 min):

\frac{80 - 60}{5} = k (70 - 20)

\frac{20}{5} = k \cdot 50

k = \frac{4}{50} = \frac{2}{25}

For the second interval (60°C to 40°C in time

t

\frac{60 - 40}{t} = k (50 - 20)

\frac{20}{t} = \frac{2}{25} \cdot 30

t = \frac{20 \times 25}{60} = \frac{25}{3} min = 500 s

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option

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