Question

It takes $10minutes$ to cool a liquid from ${61}^{o}C$ to ${59}^o$. If room temperature is ${30}^{o}C$ then time taken in cooling from ${51}^{o}C$ to ${49}^{o}C$ is

• A. $10min$
• B. $11min$
• C. $13min$
• D. $15min$

Answer 1

The correct option is D $15min$

$\frac{dQ}{dt}=k(\theta -{\theta }_0)$

${\theta }_0$ is room temperature.

$\frac{61-59}{10}=k(\frac{61+59}{2}-30)........\left(1\right)$

$\frac{51-49}{t}=k(\frac{51+49}{2}-30).........\left(2\right)$

Dividing equation 1 and 2 we get

$t=15min$