Question

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

• A. 5°C
• B. 10°C
• C. 15°C
• D. 20°C

Answer 1

The correct option is B

10°C

According to Newton's law of cooling

$\frac{{\theta }_1-{\theta }_2}{t}=K[\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0]$

In the first case , $\frac{(60-50)}{10}=K[\frac{60+50}{2}-{\theta }_0]$

1 = K ( 55 - $\theta$ ) ....... (i)

In the second case , $\frac{(50-42)}{10}=K[\frac{50+42}{2}-{\theta }_0]$

0.8 = K ( 46 - ${\theta }_0$ ) .... (ii)

Dividing (i) by (ii) , we get $\frac{1}{0.8}=\frac{55-{\theta }_0}{46-{\theta }_0}$

or 46 - ${\theta }_0=44-0.8{\theta }_0\Rightarrow ={10}^{\circ }C$