1. Given data

• Heat capacity of body: $C = 1\,\text{J}/^\circ\text{C}$ (constant)
• Initial temperature: $T_i = 100^\circ \text{C}$
• Final temperature: $T_f = 200^\circ \text{C}$

For entropy change, we must use absolute temperature:

$T_i = 100 + 273 = 373\,\text{K}, \qquad T_f = 200 + 273 = 473\,\text{K}$

2. Entropy change of the body

Entropy is a state function, so it depends only on the initial and final states, not on the path or number of reservoirs.

For a body of constant heat capacity $C$,

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

Substituting $C=1$:

$\Delta S = \ln\left(\frac{473}{373}\right)$

This is the entropy change in both cases.

3. Check the options

Numerically,

$\frac{473}{373} \approx 1.268$

so

$\Delta S = \ln(1.268) \approx 0.237$

Now compare with listed forms:

• $\ln 2 \approx 0.693$
• $2\ln 2 \approx 1.386$
• $4\ln 2 \approx 2.772$
• $8\ln 2 \approx 5.545$

None of these equals $\ln(473/373)$.

So the correct physical result is:

$\Delta S_1 = \Delta S_2 = \ln\left(\frac{473}{373}\right)$

4. Interpretation with respect to options

Since entropy change of the body is same in both processes, the pattern must be $(\text{same},\text{same})$. Among the given options, only D has equal values in both cases.

However, its value $\ln 2$ is not numerically correct. The exact correct value should be

$\left(\ln\frac{473}{373},\,\ln\frac{473}{373}\right)$

Thus the option structure suggests D, but the value stated is incorrect.

5. Final conclusion

• Derived entropy changes: $\Delta S_1 = \Delta S_2 = \ln\left(\frac{473}{373}\right)$
• Among options, the only matching qualitative choice is D.