Question

A hot body is allowed to cool. The surrounding temperature is constant at ${30}^{\circ }C$. The body takes time $t_1$ to cool from ${90}^{\circ }C$ to ${89}^{\circ }C$ and time $t_2$ to cool from ${60}^{\circ }C$ to ${59.5}^{\circ }C$. Then,

• A. $t_2=2t_1$
• B. $t_2=\frac{t_1}{2}$
• C. $t_2=4t_1$
• D. $t_2=t_1$

Answer 1

The correct option is D $t_2=t_1$

According to Newton's Law of cooling, rate of heat transfer is given by:

$\frac{dQ}{dt}=hA(T-T_o)$

Rate of heat transfer from ${90}^{o}C$ is given by:

$\frac{d{Q}_1}{dt}=60hA$

$⟹\Delta Q_1=60Ah{t}_1.........\left(i\right)$

Rate of heat transfer from ${60}^{o}C$ is given by:

$\frac{d{Q}_2}{dt}=30hA$

$⟹\Delta Q_2=30Ah{t}_2..........\left(ii\right)$

Also, $\Delta Q=ms\Delta T$

$⟹\frac{\Delta Q_2}{\Delta Q_1}=\frac{ms\times 0.5}{ms\times 1}=\mathrm{0.5..............}\left(iii\right)$

From (i), (ii) and (iii),

$t_1=t_2$