Question

A body coo Is in 7 min from 60$^{\circ}$C to 40$^{\circ}$C. What time (in min) does it take to cooI from 40$^{\circ}$C to 28$^{\circ}$C, if surrounding temperature is 10$^{\circ}$C? (Assume Newton's law of cooling)

• A. 3.5
• B. 14
• C. 7
• D. 10

Answer 1

Answer: (C)

7

Answer 2

From Newton's law of cooling, $\, \, \, \, \, \, \, \, \, \, \frac{\theta_1 -\theta_2}{t} =\bigg(\frac{\theta_1+\theta_2}{2} - \theta_0\bigg)$ Therefore, $\frac{60^{\circ}-40^{\circ}}{7} \propto \bigg(\frac{60^{\circ}+40^{\circ}}{2}-10^{\circ}\bigg) \, \, \, \, \, \, \, \, \,$ ... (i) $\, \, \, \, \, \, \, \, \frac{40^{\circ}-28^{\circ}}{t} \propto\bigg(\frac{40^{\circ}+28^{\circ}}{2}-10^{\circ}\bigg) \, \, \, \, \, \, \, \, \,$... (ii) $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, t=7s$