Question

A cup of coffee cools from ${90}^{\circ }\text{C}$ to ${80}^{\circ }\text{C}$ in $t\text{mins}$, 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 ${20}^{\circ }\text{C}$.

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

Answer 1

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

According to Newton's law of cooling,

$\frac{T_1-T_2}{t}=K[\frac{T_1+T_2}{2}-T_0]$

For $1^{st}$ cup of coffee,

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

For $2^{nd}$ cup of coffee,

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

Dividing $\left(1\right)$ by $\left(2\right)$ we have,

$\frac{t^{\prime }}{2t}=\frac{65}{50}$

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

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

Why this question? Tip: When the difference between the system temperature and surrounding temperature is not very large, we can use the approximate form of Newton's law of cooling. $\frac{T_1-T_2}{t}=K[\frac{T_1+T_2}{2}-T_0]$ Here, $T_0=$ Surrounding temperature