Question

A cup of tea cools from ${80}^{o}C$ to ${60}^{o}C$ in the first minute when the temperature of the surrounding is ${30}^{o}C$. The decrease from ${60}^{o}C$ to ${50}^{o}C$ will be next

• A. 1 min
• B. 50 sec
• C. 48 sec
• D. 30 sec

Answer 1

The correct option is C 48 sec

Newton's law of cooling gives us the equation:
$\frac{dT}{dt}=-bA(T-T_0)$
${\int }_{T_1}^{T_2}\frac{dT}{T-T_0}={\int }_0^t(-bA)dt$
i.e.
$\frac{1}{t}ln\left(\frac{T_2-T_0}{T_1-T_0}\right)=constant$
i.e. substituting we get
$\frac{1}{60}ln\left(\frac{60-30}{80-30}\right)=\frac{1}{t}ln\left(\frac{50-30}{60-30}\right)$
Solving we get t = 48seconds