Q90 / 90 · JEE Main 2021 (22 Jul Shift 1) · Physics

In $5 minutes$ , a body cools from $75 ° C$ to $65 ° C$ at room temperature of $25 ° C .$ The temperature of body at the end of next $5 minutes$ is ${}^{\circ}C .$

Correct answer: 57

Solution

By newton's law of cooling (with approximation),
$\frac{Δ T}{Δ t} = - C (T_{ag} - T_{s})$

$1^{st} \frac{- 10 ° C}{5 \min} = - C (70 ° C - 25 ° C)$

$\Rightarrow C = \frac{2}{45} \min^{- 1}$

$2^{nd} \frac{T - 65}{5 \min} = - C (\frac{T + 65}{2} - 25) = - (\frac{2}{45}) (\frac{T + 15}{2})$

$\Rightarrow 9 (T - 65) = - (T + 15)$
$\Rightarrow 10 T = 570$
$\Rightarrow T = 57 ° C$

Hence, the correct answer is 57.