To solve the problem, we need to calculate the entropy change of the body in both cases where it is heated by different numbers of reservoirs. The body has a constant heat capacity of \( C = 1 \, \text{J/°C} \), and it is being heated from an initial temperature \( T_i = 100 \, \text{°C} \) to a final temperature \( T_f = 200 \, \text{°C} \).
### Step-by-Step Solution:
1. **Calculate the Change in Temperature:**
\[
\Delta T = T_f - T_i = 200 \, \text{°C} - 100 \, \text{°C} = 100 \, \text{°C}
\]
2. **Calculate the Heat Supplied to the Body:**
The heat \( Q \) supplied to the body can be calculated using the formula:
\[
Q = C \cdot \Delta T
\]
Substituting the values:
\[
Q = 1 \, \text{J/°C} \cdot 100 \, \text{°C} = 100 \, \text{J}
\]
3. **Calculate the Entropy Change for Each Reservoir Case:**
The entropy change \( \Delta S \) for a body when it absorbs heat \( Q \) at a constant temperature \( T \) is given by:
\[
\Delta S = \frac{Q}{T}
\]
However, since the body is heated sequentially by different reservoirs, we need to consider the average temperature at which the heat is absorbed.
- **For Case (i): Two Reservoirs:**
Each reservoir supplies half of the total heat:
\[
Q_1 = Q_2 = \frac{Q}{2} = \frac{100 \, \text{J}}{2} = 50 \, \text{J}
\]
The average temperature for the first reservoir (from \( 100 \, \text{°C} \) to \( 150 \, \text{°C} \)) is:
\[
T_1 = \frac{100 + 150}{2} = 125 \, \text{°C} = 398.15 \, \text{K}
\]
The average temperature for the second reservoir (from \( 150 \, \text{°C} \) to \( 200 \, \text{°C} \)) is:
\[
T_2 = \frac{150 + 200}{2} = 175 \, \text{°C} = 448.15 \, \text{K}
\]
The total entropy change for the two reservoirs is:
\[
\Delta S_1 = \frac{50}{398.15} + \frac{50}{448.15}
\]
- **For Case (ii): Eight Reservoirs:**
Each reservoir supplies:
\[
Q_i = \frac{Q}{8} = \frac{100 \, \text{J}}{8} = 12.5 \, \text{J}
\]
The average temperatures for each reservoir will be calculated similarly, but since there are more reservoirs, the average temperature will vary more gradually. The total entropy change will be:
\[
\Delta S_2 = \sum_{i=1}^{8} \frac{Q_i}{T_i}
\]
4. **Final Calculation:**
After calculating the individual contributions for both cases, we can summarize the total entropy changes \( \Delta S_1 \) and \( \Delta S_2 \).
