A) \[100J/{}^\circ C\]
B) \[300J/{}^\circ C\]
C) \[750J/{}^\circ C\]
D) \[1500J/{}^\circ C\]
Correct Answer: D
Solution :
[d] Rate of loss of heat \[\propto \] difference in tem- perature with the surroundings. At\[50{}^\circ C,\frac{dQ}{dt}=k\left( 50-20 \right)=10\], where k = constant \[\therefore \operatorname{k}=l/3\] At an angle temperature of\[35{}^\circ C\], \[\frac{\operatorname{dQ}}{dt}=\frac{1}{3}\left( 35-20 \right)J/s=5J/s\] Heat lost in 1 minutes \[=\frac{dQ}{dt}\times 60J=5\times 60J=300J=Q\] Fall in temperature\[= 0.2{}^\circ C= AQ\]. \[\operatorname{Q}=c\Delta \theta .\]. Heat capacity \[=c=\frac{Q}{d\theta }=\frac{300J}{{{0.2}^{\operatorname{o}}}C}=1500J{{/}^{\operatorname{o}}}C.\]
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