A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in $t$ minutes, when t

Question

A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in $t$ minutes, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken by a similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at a room temperature same at $20^{\circ} \mathrm{C}$ is\

Professor PC Thomas Classes · Answer

(c) : Initial temperature $\left(T_{i}ight)=90^{\circ} \mathrm{C}$ Final temperature $\left(T_{f}ight)=80^{\circ} \mathrm{C}$\ Room temperature ( $T_{0}$ ) $=20^{\circ} \mathrm{C}$\ Let time taken be $t$ minutes.\ According to Newton's law of cooling,\Rate of cooling $\frac{d T}{d t}=K\left[\frac{\left(T_i+T_fight)}{2}-T_0ight]$
$\frac{\left(T_f-T_iight)}{t}=K\left[\frac{(90+80)}{2}-20ight]$
$\Rightarrow \frac{90-80}{t}=K[65]$

$
K=\frac{10}{65 t}
$
$\ln 2^{	ext {nd }}$ condition,
Initial temperature, $T_i=80^{\circ} \mathrm{C}$
Final temperature, $T_f=60^{\circ} \mathrm{C}$
Let time taken be $t^{\prime}$ minutes.
Then, $\frac{(80-60)}{t^{\prime}}=\frac{10}{65 t}\left[\frac{(80+60)}{2}-20ight] \frac{20}{t^{\prime}}=\ \frac{10}{65 t}(50)$
$\Rightarrow \quad t^{\prime}=\frac{13}{5} t$