A body cools from ${50}^0\mathrm{C} \mathrm{to} {49}^{\mathrm{o}}\mathrm{C} \mathrm{in} 5 \sec . \mathrm{How} \mathrm{long} \mathrm{will} \mathrm{it} \mathrm{take} \mathrm{to} \mathrm{cool} {40}^0\mathrm{C} \mathrm{to} 39.0^0\mathrm{C}?$ Assume the temperature of surroundings to be ${30}^0\mathrm{C}$ $\mathrm{and} \mathrm{Newton}\text{'}\mathrm{s} \mathrm{law} \mathrm{of} \mathrm{cooling} \mathrm{to} \mathrm{be} \mathrm{valid}:$

According to Newton's law of cooling, the ratio of cooling directly proportional to the temperature difference. When the temperature difference is halved, the rate of cooling is also halved. So time taken is 10 sec.