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

Step 1: Identify the given parameters.

The solid body has a constant heat capacity $C$, and is heated from an initial temperature $T_1$ to a final temperature $T_2$.

$$C = 1\,\text{J}/^\circ\text{C}$$ $$T_1 = 100^\circ\text{C}$$ $$T_2 = 200^\circ\text{C}$$

Step 2: Define the entropy change of the body.

Entropy is a state function, meaning its change depends only on the initial and final states of the system, and is independent of the path taken or the number of reservoirs used.

$$\Delta S = \int_{T_1}^{T_2} \frac{dQ_{\text{rev}}}{T}$$

Since $dQ_{\text{rev}} = C\,dT$:

$$\Delta S = \int_{T_1}^{T_2} \frac{C\,dT}{T}$$ $$\Delta S = C \ln\left(\frac{T_2}{T_1}\right)$$

Step 3: Calculate the entropy change for both cases.

Since the initial and final temperatures are the same in both case (i) and case (ii), the entropy change of the body is identical for both processes. Using the temperature ratio of the given values:

$$\Delta S_{(i)} = \Delta S_{(ii)} = 1 \cdot \ln\left(\frac{200}{100}\right)$$ $$\Delta S_{(i)} = \Delta S_{(ii)} = \ln 2\,\text{J}/^\circ\text{C}$$

Answer: The entropy change of the body in the two cases is $\ln 2\,\text{J}/^\circ\text{C}$ and $\ln 2\,\text{J}/^\circ\text{C}$ respectively.