Question

A solid body of constant heat capacity $1J{/}^{\circ }C$ 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 of ${100}^{\circ }C$ to a final temperature of ${200}^{\circ }C$ . Entropy change of the body in two cases respectively is :

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

Answer 1

The correct option is B $\ln 2,\ln 2$

The entropy change from State A to B is given by
$\Delta S^{\circ }={\int }_A^B\frac{dQ}{T}$...(i)
where $dQ$ is heat supplied in J to the body and temperature is in K

$T_A=273+100K=373K;T_B=273+200=473K$
$dQ=CdT$ where C is heat capacity of body and dT is increase in temperature in $C^{\circ }$

Substituting $T^{\prime }=T-273$ in equation (i) where $T^{\prime }$ is temperature in $C^{\circ }$ and $T$ is in kelvin.
$d{T}^{\prime }=dT$
integration limits:$T_A^{\prime }=373-273=100C^{\circ };T_B^{\prime }=473-273=200C^{\circ }$
$\Delta S^{\circ }={\int }_{100}^{200}\frac{Cd{T}^{\prime }}{T^{\prime }+273}=ln\left(\frac{473}{373}\right)$
As the heat transferred is same the entropy change will be same in both the cases