Question

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

• A. $\frac{5}{13}t$
• B. $\frac{13}{10}t$
• C. $\frac{13}{5}t$
• D. $\frac{10}{13}t$

Answer 1

The correct option is C $\frac{13}{5}t$

We know that, Newton's law of cooling,

$\frac{T_1-T_2}{t}=\alpha (\frac{T_1+T_2}{2}-T_0)$

So, for the first case,

$\frac{90-80}{t}=\alpha (\frac{90+80}{2}-20)...\left(1\right)$

And for the second case,

$\frac{80-60}{t^{\prime }}=\alpha (\frac{80+60}{2}-20)...\left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$, we get,

$t^{\prime }=\frac{13t}{5}$

Hence, option $\left(C\right)$ is the correct answer.