Question

A hot body is placed in cooler surrounding. When the hot body temperature ${75}^{\circ }C$, the rate of cooling is $4^{\circ }C/min$. When it is ${55}^{\circ }C$, the rate of cooling is $2^{\circ }C/min$. The temperature of the surroundings is

• A. ${20}^{\circ }C$
• B. ${25}^{\circ }C$
• C. ${30}^{\circ }C$
• D. ${35}^{\circ }C$

Answer 1

The correct option is D ${35}^{\circ }C$

We know that,

cooling rate is directly proportional to difference of temperature of body and temperature of surroundings

$\frac{d\theta }{dt}\propto ({\theta }_t-{\theta }_s)$

$\frac{d\theta }{dt}=k({\theta }_t-{\theta }_s)$ --------------------------(1)

Given,

$At{75}^{\circ }C,\frac{d\theta }{dt}=4^{\circ }C/min$

And

$At{55}^{\circ }C,\frac{d\theta }{dt}=2^{\circ }C/min$

Substitute all value in equation(1)

$4=k(75-{\theta }_s)$ ---------------------------------------(2)

And

$2=k(55-{\theta }_s)$ ----------------------------------------(3)

By Dividing both the above equation we will get

$\frac{4}{2}=\frac{k(75-{\theta }_s)}{k(55-{\theta }_s)}$

$2=\frac{(75-{\theta }_s)}{(55-{\theta }_s)}$

$2(55-{\theta }_s)=(75-{\theta }_s)$

$110-{2\theta }_s=75-{\theta }_s$

$110-75=2{\theta }_s-{\theta }_s$

${\theta }_s=35$

Hence, Temperature of surroundings is 35 degree celcious.Hence, Option (D) is correct option.