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. $50sec$.
• B. $90sec$
• C. $60sec$.
• D. $48sec$.

Answer 1

The correct option is D $48sec$.

$\frac{\Delta t}{time}=k(\frac{t_1+t_2}{2}-t_0)$

$\frac{80-60}{60}=k(\frac{80+60}{2}-30)$

$\frac{1}{3}=k\left(40\right)$

$k=\frac{1}{120}$

$\frac{60-50}{time}=\frac{1}{120}(\frac{60+50}{2}-30)$

$Time=48secs$