Question:-
A solid copper ball of mass 500 grams, when quenched in a water bath at 30°C, cool from 530°C to 430°C in 10 seconds. What will be the temperature of the ball the next 10 seconds?
Option (A)
300°C
Option (B)
320°C
Option(C)
350°C
Option(D)
Not determinable for want of sufficient data
Correct Option:
(C)
Question Solution:
This is the case of unsteady state heat conduction
Tf = fluid temperature
T0 = initial temperature
T = temperature after elapsing time ‘t’ heat transferred = change in internal energy
hA (T-Tf) = -mcp (dT/dt)
this is derived to
(q/q0) = e-(hAt/pcp)
Or (T-T¥) / (T0-T¥) = e(-hA/pcpv)t
Or (430-30) / (530-30) =0.8 = e(-hA/pcpv)t (t=10sec)
After 20 sec (2t)
(T-30)/(530-30) = e(-hA/pcpv)(2t) = [e(-hA/pcp]2
Or (T-30)/500 = (0.8)2 = 0.64
\ T = 350°C
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