A metal ball of mass `2kg` is heated means of a `40W` heater in a room at `25^(@)C`. The temperature of the ball beomes steady at `60^(@)C`.
Find the rate of loss of heat to the surrounding when the ball is at `60^(@)C`.

Options

A. `40 W`
B. `16 W`
C. `96 W`
D. `100W`

Answer

A

Explanation

(a) At steady state
heat gained per unit time = heat lost per unit time
`40 W = heat lost`
(b) from Newton's law of cooling,
`(dQ)/(dt) prop (T -T_(0)) rArr (dQ)/(dt) = k (T -T_(0))`
`rArr at 60^(@)C, k (60 -25) = 40 rArr k = (40)/(35)`
Now, `at 39^(@)C`, rate of heat loss `39^(@)C`
`=k (39 -25) =16W`
(c ) `40 = k (60 - 25)`
`(dQ)/(dt) = k(T - 25)`
and `(dT)/(dt) = ((39-25))/(2 xx 60) = (T - 25)/(t) rArr T - 25 = (7t)/(60)`
Further `(dQ)/(dt) =(7 - 25) k = (T -25)/(35) xx 40`
`Q = f ((dQ)/(dt)) dt, Q = overset(2xx60)underset(0)int (2)/(15) xx t dt`
`Q = (2)/(15 xx2) [t^(2)]_(0)^(2xx60) rArr Q = 960 J`