Question

Hot water cools from ${60}^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$ in the first $10\text{min}$ and to ${42}^{\circ }\text{C}$ in the next $10\text{min}$. The temperature of the surrounding is

• A. $5^{\circ }\text{C}$
• B. ${10}^{\circ }\text{C}$
• C. ${15}^{\circ }\text{C}$
• D. ${20}^{\circ }\text{C}$

Answer 1

The correct option is B ${10}^{\circ }\text{C}$

$\frac{{\theta }_1-{\theta }_2}{t}\propto [\frac{{\theta }_1+{\theta }_2}{2}-\theta ]$ According to Newton's law of cooling
For the first condition
$\frac{60-50}{10}\propto [\frac{60+50}{2}-\theta ]\Rightarrow 1=K[55-\theta ]...\left(i\right)$
For the second condition
$\frac{50-42}{10}\propto [\frac{50+42}{2}-\theta ]\Rightarrow 0.8=K[46-\theta ]...\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get $\theta ={10}^{\circ }\text{C}$