Question

A body colls from ${70}^{o}C$ to ${60}^{o}C$ in 10 minutes when kept in air at${30}^{o}C$. In the next ten minutes its temprature will fall

• A. Below ${50}^{o}C$
• B. To${50}^{o}C$
• C. To above ${52.5}^{o}C$
• D. To any of the above depending on its specific heat

Answer 1

Answer: (C)

To above ${52.5}^{o}C$

Using Newton's law cooling, $kt=ln\frac{T_i-T_0}{T_f-T_0}$

In first case, $k\left(10\right)=ln\frac{70-30}{60-30}=ln\left(4/3\right)$

or $k=\left(1/10\right)ln\left(4/3\right)$

In second case, $\left(1/10\right)ln\left(4/3\right)\times \left(10\right)=ln\frac{60-30}{T_f-30}$

or $ln\frac{4}{3}=ln\frac{30}{T_f-30}$

or $4T_f-120=90$

or $T_f=210/4={52.5}^{\circ }C$

Option C is the correct answer.