A cup of coffee cools from $90^{\circ}C$ to $80^{\circ}C$ in $t$ minutes, when the room temperature is $20^{\c

Question

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

Accepted Answer

Using Newton's law of cooling: $-\left(\frac{T_1 + T_2}{2} - T_s\right)K = \frac{T_1 - T_2}{\Delta t}$. For the first case: $-K\left(\frac{90+80}{2} - 20\right) = \frac{90-80}{t} \Rightarrow -K(65) = \frac{10}{t} \Rightarrow K = \frac{-2}{13t}$. For the second case: $-K\left(\frac{80+60}{2} - 20\right) = \frac{80-60}{t_1} \Rightarrow -K(50) = \frac{20}{t_1}$. Substituting $K$: $\frac{2}{13t}(50) = \frac{20}{t_1} \Rightarrow t_1 = \frac{13t}{5}$.

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