A metal ball of mass 2 kg is heated by means of a 40 W heater in a room at 25°C. The temperature of the ball becomes steady at 60°C.
(i) Find the rate of loss of heat to the surrounding when the ball is at 60°C.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) At steady state
heat gained per unit time = heat lost per unit time
40W = heat lost
(ii) from Newton's law of cooling,
∝ (T – T 0 )
⇒ 
= k (T – T 0 )
⇒ at 60º C
k (60 – 25) = 40
⇒ k = 
Now, at 39º C, rate of heat loss 39ºC = k (39 – 25) = 16 W
(iii) 40 = k (60 – 25)
= k (T – 25)
and 
= 
= 
⇒ T – 25 = 
Further 
= (T – 25) k = 
× 40
Q = 
Q = 
Q = 
⇒ Q = 960 J
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