Question

A body at ${50}^{o}C$ cools in a surroundings maintained at ${30}^{o}C$ . The temperature at which the rate of cooling is half that of the begining is :

Answer 1

By using Newton’s Law of cooling we get

$\frac{dQ}{dt}=k{\left(50-30\right)}^{\circ }$

$\frac{dQ}{dt}=k{20}^{\circ }$

The temperature at which the rate becomes half is

$\frac{k}{2}\times {20}^{\circ }=k({50}^{\circ }-T)$

$\therefore T={40}^{\circ }$

The temperature of the surrounding should be ${40}^{\circ }$