JEE Main 2023 — Thermal Properties of Matter Question with Solution

The formula to calculate the rate of cooling of the object is given by

$\frac{dT}{dt}=K\left[\frac{T_f+T_i}{2}-T_0\right] ...\left(1\right)$

For the first case, it can be written, using equation (1) that

$\begin{array}{rcl}\frac{80-60}{5}& =& K\left[\frac{80+60}{2}-20\right]\\ 4& =& 50K ...\left(2\right)\end{array}$

If $t$ is the required time for the second case, from equation (1), it can be written that

$\begin{array}{rcl}\frac{60-40}{t}& =& K\left[\frac{60+40}{2}-20\right]\\ \frac{20}{t}& =& 30K ...\left(3\right)\end{array}$

Divide equation (2) by equation (3) and solve to calculate the required time.

$\begin{array}{rcl}\frac{4}{{\displaystyle \frac{20}{t}}}& =& \frac{50K}{30K}\\ & \Rightarrow & \frac{t}{5}=\frac{5}{3}\\ & \Rightarrow & t=\frac{25}{3} \min =500 \mathrm{s}\end{array}$