Question

In a room where the temperature is ${30}^{o}C$, a body cools from ${61}^{o}C$ to ${59}^{o}C$ in 4 min. The time (in min) taken by the body to cool from ${51}^{o}C$ to ${49}^{o}C$ will be:

• A. $4$
• B. $6$
• C. $5$
• D. $8$

Answer 1

Answer: (B)

$6$

According to Newton's Law of Cooling,

$\frac{dT}{dt}\propto (T-T_0)$

where $T_0$ is the temperature of surroundings,

$⟹t\propto ln\left(\frac{T_f-T_0}{T_i-T_0}\right)$

Thus $\frac{t^{\prime }}{4min}=\frac{ln\left(\frac{49-30}{51-30}\right)}{ln\left(\frac{59-30}{61-30}\right)}$

$⟹t^{\prime }=6min$