Question

A body cools down from ${80}^{\circ }\text{C}$ to ${60}^{\circ }\text{C}$ in 10 minutes when the temperature of surroundings is ${30}^{\circ }\text{C}$. The temperature of the body after next $10$ minutes will be

• A. ${30}^{\circ }\text{C}$
• B. ${48}^{\circ }\text{C}$
• C. ${50}^{\circ }\text{C}$
• D. ${52}^{\circ }\text{C}$

Answer 1

The correct option is B ${48}^{\circ }\text{C}$

Given that,
Initial temperature of body $T_1={80}^{\circ }\text{C}$
Temperature after 10 min $T_2={60}^{\circ }\text{C}$
Temperature of surroundings $T_s={30}^{\circ }\text{C}$

Applying Newton's law of cooling,
$\text{ln}\left(\frac{T_1-T_0}{T_2-T_0}\right)=Kt$
$\Rightarrow \text{ln}\left(\frac{80-30}{60-30}\right)=K\times 10...\left(i\right)$
Let temperature of body after next $10$ minutes be $T$.
So, $\text{ln}\left(\frac{60-30}{T-30}\right)=K\times 10...\left(ii\right)$
Using equation $\left(i\right)$ and $\left(ii\right)$,
$\text{ln}\frac{5}{3}=\text{ln}\frac{30}{T-30}$
$\Rightarrow 30\times 3=5T-5\times 30$
$\Rightarrow T={48}^{\circ }\text{C}$