Properties Of Solids And Liquids Question 460
Question: A system S receives heat continuously from an electrical heater of power 10 W. The temperature of S becomes constant at $ 50{}^\circ C $ when the surrounding temperature is $ 20{}^\circ C $ . After the heater is switched off; S cools from $ 35.1{}^\circ C $ to $ 34.9{}^\circ C $ in 1 minute. The heat capacity of S is
Options:
A) $ 100J/{}^\circ C $
B) $ 300J/{}^\circ C $
C) $ 750J/{}^\circ C $
D) $ 1500J/{}^\circ C $
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Answer:
Correct Answer: D
Solution:
[d] Rate of loss of heat $ \propto $ difference in tem- perature with the surroundings.
At $ 50{}^\circ C,\frac{dQ}{dt}=k( 50-20 )=10 $ , where k = constant $ \therefore k=l/3 $
At an angle temperature of $ 35{}^\circ C $ , $ \frac{dQ}{dt}=\frac{1}{3}( 35-20 )J/s=5J/s $ .
Heat lost in 1 minutes $ =\frac{dQ}{dt}\times 60J=5\times 60J=300J=Q $
Fall in temperature $ = 0.2{}^\circ C= AQ $ . $ Q=c\Delta \theta . $
Heat capacity $ =c=\frac{Q}{d\theta }=\frac{300J}{{{0.2}^{o}}C}=1500J{{/}^{o}}C. $
BETA
