A metal ball of mass `1kg` is heated by means of a `20W` heater in a room at `20^(@)C` . The temperature of the ball becomes steady at `50^(@)C` . (a) Find the rate of loss of heat to the surrounding when the ball is at `50^(@)C` . (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at `30^(@)C` . (c) Assume that the temperature of the ball rises uniformly from `20^(@)C` to `30^(@)C` in `5` minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal.

Explanation

Rate of loss of heat = power of heater `= 20 W`
(b) `(dQ)/(dt) = - k_(0) (theta - theta_(0))`
`= 20 = - k_(0) (50 - 20) implies k_(0) = - (2)/(3)`
`(dQ)/(dt) = - k_(0) (theta - theta_(0)) = (2)/(3) (30 - 0)`
`= (20)/(3) W`
(c ) Temperature of ball as a function of time `t` s
`theta = 20 + (30 - 20)/(5 xx 60) t = 20 + (t)/(30)`
`(dQ)/(dt) = - k (theta - theta_(0)) = (2)/(3) [20 + (t)/(30) - 20]`
` = (t)/(45)`
`int dQ = int_(theta)^(300) (t)/(5) dt = (1)/(45). |(t^(2))/(2)|_(0)^(300) = (1)/(90) xx 300 xx 300`
`Q = 1000 J`