Why heating in small steps generate less entropy

Question

I was learning about entropy the other day and did some exercises. This thing I didn't quite understand. For example, if we heat a bowl of water 50C from 20 to 70 it generates way more entropy than if we calculate heating using 10C steps. Why is that?

Answer 1

Assume that the heat capacity of water $C_w$ is constant and denote the equilibrium temperatures as $T_k$ where $k=0,1,..M$ are the steps you take to reach the $T_f$ from the initial $T_0$ temperature. Then at each step above the equilibrium the infinitesimal entropy change is $$dS = C_w\left ( \frac{1}{T} - \frac{1}{T_{k}}\right) dT \tag{1}$$ Let us sum up these to get the total entropy change: $$\Delta S = \sum_{k=1}^{M} \int_{T_{k-1}}^{T_{k}}dS_k = \sum_{k=1}^{M}\int_{T_{k-1}}^{T_{k}}C_w\left ( \frac{1}{T} - \frac{1}{T_k}\right) dT$$

$$= C_w\int_{T_0}^{T_M} \frac{1}{T} dT - C_w\sum_{k=1}^{M}\int_{T_{{k-1}}}^{T_{k}}\left ( \frac{1}{T_k}\right) dT$$

$$= C_w \left(ln \frac{T_f}{T_0} - \sum_{k=1}^{M}\left ( \frac{T_{k}-T_{k-1}}{T_k}\right) \right)$$

$$= C_w \left(ln \frac{T_f}{T_0} - M +\sum_{k=1}^{M}\frac{T_{k-1}}{T_k} \right)$$ Now we have $M$ temperatures and we wish to minimize the sum $\sum_{k=1}^{M}\frac{T_{k-1}}{T_k}$ for $T_0<T_1<...<T_{M-1}<T_{M}=T_f$ for given $T_0,T_f$ and $M$. Using the inequality of the arithmetic and geometric means (https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means) we see that $$\frac{1}{M}\sum_{k=1}^{M}\frac{T_{k-1}}{T_k} \ge \left(\prod_{k=1}^{M}\frac{T_{k-1}}{T_k} \right)^{1/M}=\left(\frac{T_0}{T_M} \right)^{1/M}$$

with equality iff for all $k$ $$\frac{T_{k}}{T_{k+1}}=\frac{T_{k-1}}{T_k}$$ from which we get successively that $T_2 = \frac{T_1^2}{T_0}$, $T_3 = \frac{T_2^2}{T_1} = \frac{T_1^3}{T_0^2}$, $T_4 = \frac{T_3^2}{T_2} = \frac{T_1^4}{T_0^3}$, and in general $T_k = \frac{T_1^k}{T_0^{k-1}}$. For$k=M$ we have $T_M = \frac{T_1^M}{T_0^{M-1}}$ from which we can express $T_1$ as $T_1 = T_0\left(\frac{T_M}{T_0}\right) ^{1/M}$ and thus the optimum temperature settings are $$T_k=T_0 \left( \frac{T_M}{T_0}\right)^{k/M} \tag 2$$ Using $T_f=T_M$ the corresponding entropy change is $$\Delta S \ge C_w \left(ln\frac{T_f}{T_0}-M+M\left(\frac{T_0}{T_f}\right)^{1/M} \right) \tag 3$$ Next for fixed $x$ and large $M$ expand $ln(x)-M+Mx^{-1/M}$ as $$ln(x)-M+Mx^{-1/M} =ln(x) - M + exp(-ln(x)/M)$$ $$\approx ln(x)-M+M \left( 1-\frac{ln(x)}{M} + \frac{1}{2} \frac{ln(x)^2}{M^2}\right)+....$$ $$\approx ln(x)-M + M - ln(x) + \frac{1}{2} \frac{ln(x)^2}{M}+....$$ $$\approx \frac{1}{2} \frac{ln(x)^2}{M}+....$$ Use this in (3) $$\Delta S \ge \frac{C_w}{2M} \left(ln\frac{T_f}{T_0}\right)^2 + ...$$ with the lower limit (equality) for the optimum steps of (2). As $M \rightarrow \infty$ the lower limit is $0$.

Answer 2

The rate at which entropy is generated (in heating situations like this) is proportional to the square of the local temperature gradient, integrated over the volume of the material. If the material is subjected to small temperature steps (say, by contact with reservoirs with different temperatures), and, during each step, it is allowed to equilibrate, the temperature gradients during each step are much lower, and the amount of entropy generated is much less. In either case, the entropy change of the water is independent of the heating history. So, although the entropy is generated within the water, less entropy gets transferred in from the single reservoir in the single step situation than from the series of reservoirs (at different temperatures) in the multi step situation. So the decrease in entropy of the single reservoir is less than the decrease in entropy of the sequence of reservoirs at the different temperatures. The total change in entropy for the combination of water and reservoirs is thus greater in the single reservoir case.

ADDENDUM

Consider the case of a rod of length L. The rod is initially at temperature $T_0$, and, at time t = 0, the end at x = 0 is suddenly changed to $T_1$ while the end at x = L continues insulated. The transient heat conduction for the rod is given by: $$\rho C\frac{\partial T}{\partial t}=k\frac{\partial^2T}{\partial x^2}$$where $\rho$ is the density, C is the heat capacity, and k is the thermal conductivity. If we multiply this equation by the cross sectional area of the rod, A, and integrate between x = 0 and x =L, we obtain: $$\frac{d\left[\int_0^L{\rho CA (T-T_0)dx}\right]}{dt}=-kA\left[\frac{\partial T}{\partial x}\right]_{x=0}$$But, $$\frac{d\left[\int_0^L{\rho CA (T-T_0)dx}\right]}{dt}=\frac{d(\Delta U)}{dt}$$and $$-kA\left[\frac{\partial T}{\partial x}\right]_{x=0}=\frac{dQ}{dt}$$where $\Delta U$ is the change in internal energy between time 0 and time t, and Q is the cumulative heat transferred at x = 0 up to time t. So, we have: $$\frac{d(\Delta U)}{dt}=\frac{dQ}{dt}$$This is nothing more than a transient form of the first law of thermodynamics for the rod.

Next let's consider what we get if we multiply the transient heat conduction equation by 1/T, and integrate between x = 0 and x=L: $$\frac{d\left[\int_0^L{\rho CA \ln{\frac{T}{T_0}}dx}\right]}{dt}=kA\int_0^L{\left[\frac{1}{T}\frac{\partial^2T}{\partial x^2}\right]dx}$$If we integrate the right hand side by parts, we obtain: $$\frac{d\left[\int_0^L{\rho CA \ln{\frac{T}{T_0}}dx}\right]}{dt}=-\frac{kA}{T_1}\left[\frac{\partial T}{\partial x}\right]_{x=0}+\int_0^L{\frac{kA}{T^2}\left(\frac{dT}{dx}\right)^2}dx$$But, $$\frac{d\left[\int_0^L{\rho CA \ln{\frac{T}{T_0}}dx}\right]}{dt}=\frac{d(\Delta S)}{dt}$$and $$-\frac{kA}{T_1}\left[\frac{\partial T}{\partial x}\right]_{x=0}=\frac{d(Q/T_1)}{dt}$$where $\Delta S$ is the change in entropy of the rod between time zero and time t and $Q/T_1$ is the cumulative amount of heat added at the boundary x = 0 divided by the (constant) boundary temperature $T_1$. So we have, $$\frac{d(\Delta S)}{dt}=\frac{d(Q/T_1)}{dt}+\int_0^L{\frac{kA}{T^2}\left(\frac{dT}{dx}\right)^2}dx$$ If we integrate this equation with respect to time, we obtain:$$\Delta S=Q/T_1+\int_0^t{\int_0^L{\frac{kA}{T^2}\left(\frac{dT}{dx}\right)^2dx}dt}\tag{1}$$Note that the second term on the right hand side of this equation is positive definite. Therefore, it follows that $$\Delta S\geq Q/T_1$$This is the Clausius inequality for the system under consideration. It also follows from Eqn. 1 that the second term on the right hand side represents physically the cumulative amount of entropy generated in the rod between time t = 0 and time t (while the first term on the right hand side represents the amount of entropy transferred to the rod from the reservoir at temperature $T_1$ at the boundary x = 0).