A metal ball of 1kg mass is heated by a 20W heater in a room at`20^(@)C`. After some time temperature of ball becomes steady at `50'^(@)C`. Find the rate of loss of heat by the ball to surrounding when its temperature becomes `50^(@)C`. Also find the rate at which it looses heat to the surrounding when its temperature was `30^(@)C`.

Explanation

It is given that when ball is at `50^(@)C`, its temperature becomes steady. This means that rate at which heat is being supplied to the ball, at the same rate, it is loosing heat. Thus when ballis at `50^(@)C`, the rate of loss of heat to the surrounding by ball is
`(dQ)/(dt) =Power of Heater`
=20W
When ball was at `30^(@)C`, using Newton's Law of cooling we can state that the sate of heat loss by ball is directly proportional to its temperature difference with surrounding, thus we have
`(dQ)/(dt) =k(30-20)`
`(dQ)/(dt) =10k`...(4.61)
From the state when ball was at `50^(@)C`, we can write from Newton's Law of cooling
`20 =k(50-20)`
or `k=(2)/(3)`
Thus from equation-(4.61), we can write
`(dQ)/(dt) = 10 xx (2)/(3)=(20)/(3)W`.