Question

A solid body of constant heat capacity $1 \,J/?C$ is being heated by keeping it in contact with reservoirs in two ways : (i) Sequentially keeping in contact with $2$ reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with $8$ reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature $100^{\circ}C$ to final temperature $200^{\circ}C$. Entropy change of the body in the two cases respectively is :

• A. ln 2, 2 ln 2
• B. 2 ln 2, 8 ln 2
• C. ln 2, 4 ln 2
• D. ln 2, ln 2

Answer 1

The Correct Option is D

The entropy change from State $A$ to $B$ is given by
$\Delta S ^{\circ}=\int\limits_{ A }^{ B } \frac{ d Q }{ T } \ldots$ (i)
where $dQ$ is heat supplied in $J$ to the body and temperature is in $K$
$T _{ A }=273+100 \,K =373 \,K$;
$T _{ B }=273+200=473 \,K$
$dQ = C dT$
where $C$ is heat capacity of body and $dT$ is increase in temperature in $C ^{\circ}$
Substituting $T'= T -273$ in equation (i)
where $T'$ is temperature in $C ^{\circ}$ and $T$ is in kelvin.
$dT'= dT$
integration limits: $T _{ A }'=373-273=100 C ^{\circ} $
$; T _{ B }'=473-273=200 \,C ^{\circ}$
$\Delta S^{\circ}=\int\limits_{100}^{200} \frac{C d T'}{T'+273}=\ln \left(\frac{473}{373}\right)$
As the heat transferred is same the entropy change will be same in both the cases