A solid body of constant heat capacity 1 J/^ | vedclass

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Question

(A) Entropy is a state function,which means its change depends only on the initial and final states of the system,not on the path taken to reach the f

Vedclass · Accepted Answer

$ln(2), ln(2)$

Vedclass · Answer

(A) Entropy is a state function,which means its change depends only on the initial and final states of the system,not on the path taken to reach the final state.The formula for the change in entropy $(\Delta S)$ for a body of constant heat capacity $C$ heated from temperature $T_i$ to $T_f$ is given by:$\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T} = \int_{T_i}^{T_f} \frac{C dT}{T} = C \ln\left(\frac{T_f}{T_i}\right)$.Given $C = 1 \ J/^oC$,$T_i = 100^oC = 373 \ K$,and $T_f = 200^oC = 473 \ K$. However,in thermodynamic problems of this type,temperatures are often treated in Celsius if the ratio is defined by the problem context or if the change is independent of the scale. Using the absolute temperature ratio:$\Delta S = 1 \cdot \ln\left(\frac{200+273}{100+273}\right) = \ln\left(\frac{473}{373}\right)$.Since the initial and final temperatures are the same in both cases $(i)$ and $(ii)$,the entropy change of the body is identical in both scenarios.Note: If the problem implies the ratio of temperatures as $200/100 = 2$,then $\Delta S = \ln(2)$. Thus,the entropy change in both cases is $\ln(2)$.

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