Question

A cup of tea cools from ${80}^{o}C$ to ${60}^{o}C$ in one minute. The ambient temperature is ${30}^{o}C$. In cooling from ${60}^{o}C$ to ${50}^{o}C$ it will take

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\frac{80-60}{1}=K(\frac{80+60}{2}-30)\Rightarrow K=\frac{1}{2}$
Again $\frac{60-50}{t}=\frac{1}{2}(\frac{60+50}{2}-30)\Rightarrow$
$t=0.8\times 60=48sec$