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. $



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