Heating a mass of water: one, two, then $N$ thermal reservoirs

Question 1. The diagram shows the water initially at $T_i$, the reservoir at $T_f$ and a heat-flow arrow from the reservoir to the water. A boundary encloses the isolated combined system $S+\text{surroundings}$. The water reaches $T_f$ without receiving work, so the first law gives

$$Q_{\mathrm{water}}=\Delta U_{\mathrm{water}}=mc(T_f-T_i)>0.$$

The combined water—reservoir system is isolated: its total energy is conserved,

$$\Delta U_{\mathrm{water}}+\Delta U_{\mathrm{th}}=0.$$

Since neither body exchanges work,

$$Q_{\mathrm{water}}+Q_{\mathrm{th}}=0,\qquad Q_{\mathrm{th}}=-Q_{\mathrm{water}}=-mc(T_f-T_i)<0.$$

The heat received by the reservoir is negative: it transfers energy to the water.

Question 2. Along a reversible path from $T_i$ to $T_f$, the water receives $\delta Q_{\mathrm{rev}}=mc\,\mathrm dT$. Its entropy change is

$$\Delta S_{\mathrm{water}}=mc\int_{T_i}^{T_f}\frac{\mathrm dT}{T} =mc[\ln T]_{T_i}^{T_f}=mc\ln\frac{T_f}{T_i}>0.$$

The reservoir remains at $T_f$, hence

$$\Delta S_{\mathrm{th}}=\frac{Q_{\mathrm{th}}}{T_f} =-mc\frac{T_f-T_i}{T_f}<0.$$

Since the combined system is isolated, it exchanges no entropy with the rest of the universe. Its entire entropy change is therefore produced internally:

$$\Delta S_{\mathrm{tot}}^{(1)} =\Delta S_{\mathrm{water}}+\Delta S_{\mathrm{th}} =mc\left[\ln\frac{T_f}{T_i}-\frac{T_f-T_i}{T_f}\right]>0.$$

Indeed, $1/T>1/T_f$ for $T_i\leq T<T_f$, so $\int_{T_i}^{T_f}\mathrm dT/T>(T_f-T_i)/T_f$. Heat transfer between bodies at different temperatures is irreversible.

Question 3. Set $h=(T_f-T_i)/2$, so that $T_m-T_i=T_f-T_m=h$. The water receives $mc h$ at each step. Isolation of the combined system gives, for the reservoirs,

$$Q_{\mathrm{th},1}=-mc h,\qquad Q_{\mathrm{th},2}=-mc h,$$

and therefore

$$\Delta S_{\mathrm{th},1}=-\frac{mc h}{T_m},\qquad \Delta S_{\mathrm{th},2}=-\frac{mc h}{T_f}.$$

For the water, add the changes over the two steps:

$$\begin{aligned} \Delta S_{\mathrm{water}} &=mc\ln\frac{T_m}{T_i}+mc\ln\frac{T_f}{T_m}\\ &=mc\bigl[(\ln T_m-\ln T_i)+(\ln T_f-\ln T_m)\bigr]\\ &=mc\ln\frac{T_f}{T_i}. \end{aligned}$$

The intermediate terms cancel. The water's entropy change is exactly the same as with one reservoir: entropy is a state function, and the water's initial and final states are identical in both experiments. Only the path changes. For the isolated combined system,

$$\Delta S_{\mathrm{tot}}^{(2)} =mc\left[\ln\frac{T_f}{T_i}-h\left(\frac1{T_m}+\frac1{T_f}\right)\right].$$

The difference between the two balances is

$$\Delta S_{\mathrm{tot}}^{(1)}-\Delta S_{\mathrm{tot}}^{(2)} =mc h\left(\frac1{T_m}-\frac1{T_f}\right)>0.$$

Each contact still produces entropy, but the intermediate step reduces the total entropy produced.

Question 4. Set $T_0=T_i$ and $T_N=T_f$. At step $k$, the water goes from $T_{k-1}$ to $T_k$ and receives $mc h$. The corresponding reservoir remains at $T_k$ and receives

$$Q_{\mathrm{th},k}=-mc h,\qquad \Delta S_{\mathrm{th},k}=-\frac{mc h}{T_k}.$$

For the water, the calculation telescopes again:

$$\begin{aligned} \Delta S_{\mathrm{water}} &=mc\sum_{k=1}^{N}\ln\frac{T_k}{T_{k-1}}\\ &=mc\left[\ln\frac{T_1}{T_0}+\ln\frac{T_2}{T_1} +\cdots+\ln\frac{T_N}{T_{N-1}}\right]\\ &=mc\ln\left(\frac{T_1}{T_0}\frac{T_2}{T_1} \cdots\frac{T_N}{T_{N-1}}\right)\\ &=mc\ln\frac{T_N}{T_0} =mc\ln\frac{T_f}{T_i}. \end{aligned}$$

Each intermediate temperature appears once in a numerator and once in a denominator. Increasing the number of reservoirs still does not change the water's entropy change between the same initial and final states. However, the entropy lost by the reservoirs depends on $N$:

$$\Delta S_{\mathrm{th}}^{(N)}=-mc\,h\sum_{k=1}^{N}\frac1{T_k},\qquad \Delta S_{\mathrm{tot}}^{(N)} =mc\left[\ln\frac{T_f}{T_i}-h\sum_{k=1}^{N}\frac1{T_k}\right].$$

As $N\to\infty$, $h\to0$ and the Riemann sum tends to

$$h\sum_{k=1}^{N}\frac1{T_k} \longrightarrow\int_{T_i}^{T_f}\frac{\mathrm dT}{T} =\ln\frac{T_f}{T_i}.$$

Thus, the entropy produced in the isolated combined system tends to

$$\lim_{N\to\infty}\Delta S_{\mathrm{tot}}^{(N)} =mc\left[\ln\frac{T_f}{T_i}-\ln\frac{T_f}{T_i}\right]=0.$$

The water always gains the same entropy, while in the limit the reservoirs lose exactly that amount. Reversible heating is the ideal limit of contacts made arbitrarily slowly with an infinitesimal temperature difference.