A solid body of constant heat capacity 1 J ($$^\circ$$C)$$^{-1}$$ 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 the same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both cases the body is brought from initial temperature 100 K to final temperature 200 K. Entropy change of the body in the two cases respectively is: Note: This question was awarded as a bonus since temperatures were given in centigrade instead of in Kelvin. Proper corrections are made in the question to avoid it.

We have a solid whose heat capacity is constant and equal to $$C = 1\ \text{J K}^{-1}$$. The body is to be warmed from an initial temperature $$T_i = 100\ \text{K}$$ to a final temperature $$T_f = 200\ \text{K}$$. The total heat that must be supplied is therefore obtained from the basic relation $$Q = C\,(T_f - T_i)$$. Substituting the given numbers,

$$Q_{$$ total $$} = 1\,(200 - 100) = 100\$$ J $$.$$

For the entropy change of the body we shall repeatedly use the fundamental formula for a system with constant heat capacity,

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

Because $$C = 1$$, this simplifies to

$$\Delta S = \ln\!\left(\frac{T_f}{T_i}\right).$$

Notice carefully that entropy is a state function; it depends only on the initial and final equilibrium states of the body, not on the particular sequence of reservoirs used to raise the temperature. Nevertheless, we shall still examine both procedures in detail so that every algebraic step is explicit.

Procedure (i): two reservoirs supplying equal heat

Each reservoir contributes

$$Q_1 = Q_2 = \frac{Q_{$$ total $$}}{2} = \frac{100}{2} = 50\$$ J $$.$$

Since $$C = 1\ \text{J K}^{-1}$$, a heat of 50 J corresponds to a temperature rise of 50 K in each step. Hence after the first reservoir the temperature is

$$T_1 = 100\$$ K $$+ 50\$$ K $$= 150\$$ K $$,$$

and after the second reservoir it reaches the desired $$200\ \text{K}$$. The entropy change of the body over the whole process is

$$\Delta S_{$$ (i) $$} = \ln\!\left(\frac{200}{100}\right) = \ln 2.$$

Procedure (ii): eight reservoirs supplying equal heat

Now each reservoir supplies

$$Q_{$$ each $$} = \frac{Q_{$$ total $$}}{8} = \frac{100}{8} = 12.5\$$ J $$.$$

With the same heat capacity, the temperature rise per step is

$$\Delta T = 12.5\ \text{K}.$$

Thus the temperature sequence becomes $$100,\ 112.5,\ 125,\ 137.5,\ 150,\ 162.5,\ 175,\ 187.5,\ 200\$$ K. Irrespective of this finer partition, the initial and final temperatures remain exactly the same as before, so the entropy change of the body is again

$$\Delta S_{$$ (ii) $$} = \ln\!\left(\frac{200}{100}\right) = \ln 2.$$

Putting the two results together we have

$$\Delta S_{$$ (i) $$} = \ln 2,\qquad \Delta S_{$$ (ii) $$} = \ln 2.$$

Hence, the correct answer is Option C.