Question

A body cools down from $50{}^{\circ }\text{C}$ to $45{}^{\circ }\text{C}$ in $5\text{minutes}$ and to $40{}^{\circ }\text{C}$ in another $8\text{minutes}$. Find the temperature of the surrounding.

• A. $32{}^{\circ }\text{C}$
• B. $34{}^{\circ }\text{C}$
• C. $36{}^{\circ }\text{C}$
• D. $38{}^{\circ }\text{C}$

Answer 1

The correct option is B $34{}^{\circ }\text{C}$

Let the surrounding temperature be $T_0$.

Case 1: The mean temperature of the body when it cools down from $50{}^{\circ }\text{C}$ to $45{}^{\circ }\text{C}$ in $5\text{minutes}$ is,

$T=\frac{{50}^{\circ }+45}{2}=47.5{}^{\circ }\text{C}$

Using Newton's law of cooling,

$\frac{dT}{dt}=-k(T-T_0)$

$\Rightarrow \frac{50-45}{5}=-k(47.5-T_0)$

$\Rightarrow 1=-k(47.5-T_0)...\left(i\right)$

Case 2 : The mean temperature of the body when it cools down from $45{}^{\circ }\text{C}$ to $40{}^{\circ }\text{C}$ in $8$ minutes is,

$T=\frac{45+40}{2}=42.5{}^{\circ }\text{C}$

So, using Newton's law of cooling

$\frac{45-40}{8}=-k(42.5-T_0)$

$\Rightarrow \frac{5}{8}=-k(42.5-T_0)....\left(ii\right)$

Dividing equation $\left(i\right)$ by $\left(ii\right)$,

$\frac{8}{5}=\frac{47.5-T_0}{42.5-T_0}$

$\Rightarrow 340-8T_0=237.5-5T_0$

$\therefore T_0\simeq 34{}^{\circ }\text{C}$

Hence, $\left(\text{B}\right)$ is the correct answer.

Why this question? Since the given situation happening at room temperature. So, in this case, the temperature difference between body and surrounding will be small compared to surrounding. Hence, we can use newton's law of cooling.