Comparison of entropy change in a system and in a heat reservoir

Question

Lets consider a thermodynamic system at $T_1$ and a heat reservoir at $T_0$, where $T_1>T_0$.

The system cools down, i.e. it gives of heat to the heat reservoir (the heat reservoir can be thought of as the environment). The system, as it looses heat, has a negative entropy change. On the other hand, the heat reservoir is being heated up and its entropy increases, i.e. it has a positive entropy change.

By the second law, the following inequality must hold:

$$ \Delta S_{\space isolated\space system} = \Delta S_{\space system} + \Delta S_{\space heat\space reservoir} > 0 \tag1.$$

If the amount of heat that the system gave out is in absolute values equal (but opposite in sign) to the heat that the heat reservoir accepted, how come than that the entropy change of the heat reservoir $\Delta S_{\space heat\space reservoir}$ is still larger that the $\Delta S_{\space system}$ in absolute values, so that the equation $(1)$ would always produces a positive value?

Why if the heat (absolute) value is the same, wouldn't the $\Delta S_{\space system}$ and $\Delta S_{\space heat\space reservoir}$ simply cancel out and the change of the isolated system would than be equal to zero?

What causes the heat reservoir to have (in absolute values) a greater entropy change than the one the system has?

Answer 1

It's the same principle governing the heat $\Delta Q_{01}$ exchanged between system $0$ and system $1$, taken as positive if it increases the energy of system $0$ (badly speaking "$\Delta Q_{01} > 0$ means that heat flows from $1$ to $0$").

Heat transfer between system $0$ and system $1$ is governed by the inequality

\begin{equation} \Delta Q_{01} \left(\dfrac{1}{T_0} - \dfrac{1}{T_1} \right) \ge 0 \ , \end{equation}

i.e. "heat flows from $1$ to $0$" (very very bad to say, everytime one says it) if $T_1 > T_0$. Since it's possible to define the heat exchange in the opposite direction $\Delta Q_{10} = - \Delta Q_{01}$, it's possible to write

\begin{equation} \dfrac{\Delta Q_{01}}{T_0} + \dfrac{\Delta Q_{10}}{T_1} \ge 0 \ , \end{equation}

that should resemble at your eyes a similar inequality involving entropy of the two systems.

Answer 2

To answer the question why $$\Delta S_{\space heat\space reservoir} > \Delta S_{\space system}$$

we have to go to the definition of entropy:

$$dS=\frac{\delta Q}{T}.$$

Now for the system:

$$dS_{\space system}=\frac{\delta Q}{T_1(t)}$$

where $T_1(t)$ is a function of time, in a way that it decreases as time (and heat exchange) passes by.

For the environement:

$$dS_{\space heat\space reservoir}=\frac{\delta Q}{T_0}$$

where $T_0$ is constant temperature of the heat reservoir and is for every $t$ until the system and the heat reservoir come into thermal equilibrium lower than $T_1$.

Now to obtain the true $\Delta S_{\space system}$ we would have to know the function $T_1(t)$, but actually we don't have to, we just have to know it is bigger than $T_0$ for all $t$ (because at exactly those conditions heat exchange would take place anyway).

As basics stated, we will use the formula:

$$ \Delta S_{\space isolated\space system}= \dfrac{\Delta Q_{01}}{T_0} + \dfrac{\Delta Q_{10}}{T_1(t)} \ge 0. \ $$

Since $$T_0 < T_1(t)\space \forall\space t $$ we get that $$\dfrac{\Delta Q_{01}}{T_0}>\dfrac{\Delta Q_{10}}{T_1(t)}$$

and so:

$$ \Delta S_{\space isolated\space system}>0$$

as expected.

Answer 3

What you call "cool down" is a workless entropy transport, that is conduction. Each energetic exchange form, be it thermal, mechanical, electric, gravitational, magnetic, chemical, etc., is associated with an extensive quantity $X_k$ and its conjugate potential $Y_k$. These are entropy - temperature, volume - pressure, electric charge - electric potential, gravitational mass - gravitational potential, etc., so that the work exchanged between the system and its environment is $$\sum_k \delta w_k = \sum_k X_k \delta Y_k\\ =S\delta T - V\delta p + e\delta \psi + m\delta \phi + ...\tag{1}$$ where the extensive quantity of $X_k$ is transported between two intensive levels separated by $\delta Y_k$.

In a reversible process in which a thermodynamic system is in energy exchange with its environment the total work is conserved, that is the total work sum is zero, $$\sum_k \delta w_k = \sum_k X_k\delta Y_k = 0 \tag{2}.$$ In an irreversible process, the total work, between the system and its environment, is positive and equal to the dissipated work, call it the evolved heat, and is equal to the generated entropy $dS^* $ multiplied by the temperature $T^*$ at which it has been generated: $$\sum_k \delta w_k = \sum_k X_k \delta Y_k = T^* dS^* \tag{3}$$

What we call heat conduction is an energy exchange between two bodies in which the energy transfer involves only an entropy transport between them. Per force that must be an irreversible process because it takes at least two energetic processes to balance to zero in (1). As result it can only take the form of $S\delta T = T^*\delta S^*$. This means that if we "drop" $S$ entropy from a higher $T_h$ to a lower $T_{\ell}$ temperature separated by $\delta T=T_h-T_{\ell}$ then then there will be generated $\delta S^*$ entropy. If $\delta T$ is an infinitesimal then $T^*$ is just $T_{\ell}\approx T^*$. In this case going from the higher to the lower grade, we can also write $\delta T=-dT$, and then $-SdT = TdS$ or equivalently $$d(TS)=0\tag{4}$$ expressing conservation of thermal energy not entropy in heat conduction, lower the temperature in the product the more entropy is being transported further down.

This (4) is not a law of thermodynamics, this is a definition of a particular irreversible process in which only one type of work, namely thermal work takes place between bodies, all other type of energetic interactions are ignored. It is not a law but rather a constitutive relationship, it defines a type of material that connects the two bodies. It is akin to the magnetic constitutive relationship $\mathbf B =\mu \mathbf H$ and not to the Maxwell's equation $\mathbf {curl H} = \mathbf J$.