Question

If as liquid takes $30s$ in cooling from ${95}^{\circ }C$ to ${90}^{\circ }C$ and $70s$ in cooling from ${55}^{\circ }C$ to ${50}^{\circ }C$ then temperature of room is:-

• A. ${16.5}^{\circ }C$
• B. ${22.5}^{\circ }C$
• C. ${28.5}^{\circ }C$
• D. ${32.5}^{\circ }C$

Answer 1

The correct option is B ${22.5}^{\circ }C$

We know that,

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

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

$dt=30s$

$d\theta =95-90=5^{\circ }C$

${\theta }_t=\frac{95+90}{2}={92.5}^{\circ }C$

Substitute above value in equation (1) we will get

$\frac{5}{30}=k(92.5-{\theta }_s)$-----------------------------------(2)

Similarly,

$dt=70s$

$d\theta =55-50=5^{\circ }C$

${\theta }_t=\frac{55+50}{2}={52.5}^{\circ }C$

Substitute above value in equation (1) we will get

$\frac{5}{70}=k(52.5-{\theta }_s)$-----------------------------------(2)

By Dividing both the above equation we will get

$\frac{70}{30}=\frac{k(92.5-{\theta }_s)}{k(52.5-{\theta }_s)}$

$7(52.5-{\theta }_s)=3(92.5-{\theta }_s)$

$367.5-{7\theta }_s=277.5-{3\theta }_s$

$367.5-277.5=7{\theta }_s-{3\theta }_s$

${4\theta }_s=90$

${\theta }_s=22.5$

Hence, Temperature of surroundings is 22.5 degree celcious.Hence, Option (B) is correct option.