Question

A cup of tea cools from ${80}^{\circ }C$ to ${60}^{\circ }C$ in one minute.

The ambient temperature is ${30}^{\circ }C$.

In coolingfrom ${60}^{\circ }C$ to ${50}^{\circ }Citwilltakenearly$

• A. 30 seconds
• B. 60 seconds
• C. 90 seconds
• D. 50 seconds

Answer 1

The correct option is D

50 seconds

$\frac{80-60}{1}=K(\frac{80+60}{2}-30)\Rightarrow k=\frac{1}{2}$

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