Question

A body cools from 67C to 37C. If this takes time t when the surrounding temperature is 27C,
what will be the time taken if the surrounding temperature is 7C?
A. 2t
B. t/3
C. t/2
D. t/4

Answer 1

$$\begin{gathered} \mathrm{Case}\mathrm{I}: \\ \mathrm{From}\mathrm{Newton}\text{'}\mathrm{s}\mathrm{law}\mathrm{of}\mathrm{cooling}, \\ \frac{\mathrm{d\theta }}{\mathrm{dt}}=-\mathrm{k}\left[\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right] \\ \frac{67-37}{\mathrm{t}}=-\mathrm{k}\left[\frac{67+37}{2}-27\right] \\ \frac{30}{\mathrm{t}}=-25\mathrm{k} \\ \mathrm{k}=-\frac{30}{25\mathrm{t}} \\ \mathrm{Case}\mathrm{II}: \\ \mathrm{From}\mathrm{Newton}\text{'}\mathrm{s}\mathrm{law}\mathrm{of}\mathrm{cooling}, \\ \frac{\mathrm{d\theta }}{\mathrm{dt}}=-\mathrm{k}\left[\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right] \\ \frac{67-37}{\mathrm{t}\text{'}}=-\mathrm{k}\left[\frac{67+37}{2}-7\right] \\ \frac{30}{\mathrm{t}\text{'}}=-45\mathrm{k} \\ \mathrm{t}\text{'}=-\frac{30}{45\mathrm{k}} \\ =-\frac{30}{45\left(-\frac{30}{25\mathrm{t}}\right)} \\ =\frac{30\times 25\mathrm{t}}{45\times 30} \\ \mathbf{}\mathbf{\approx }\frac{\mathbf{t}}{\mathbf{2}} \\ \mathbf{Ans}\mathbf{:} \\ \mathbf{Option}\mathbf{\left(}\mathbf{C}\mathbf{\right)} \end{gathered}$$