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

From Newton's Law of cooling,

$\frac{\Delta T}{\Delta t}=k\left(T-T_s\right)$

where,

$\Rightarrow \frac{T_2-T_1}{t}-k\left(\frac{T_1+T_2}{2}-T_{\mathrm{s}}\right)T_{\mathrm{s}}$

Where, $T_1=$ Initial temperature of body

$T_2=$ Final temperature of body

$T_s=$ surrounding temperature

$t=$ time taken

For case$1,$

$\frac{90-80}{t}=k\left(\frac{90+80}{2}-20\right)$

$\Rightarrow \frac{10}{t}=65k$

$\Rightarrow k=\frac{10}{65t} \dots \left(\mathrm{i}\right)$

For case $2 ,$

$\frac{80-60}{t^{\text{'}}}=k\left(\frac{80+60}{2}-20\right)$

$\Rightarrow \frac{20}{t^{\text{'}}}=50\mathrm{k}$

Put the value $\mathrm{k}$ from eq.$\left(\text{i}\right)$,

$\Rightarrow \frac{20}{t^{\text{'}}}=50\times \frac{10}{65t}$

$\Rightarrow t^{\text{'}}=\frac{20\times 65}{500}t$

$\Rightarrow t^{\text{'}}=\frac{13}{5}t$