Entropy Production Due to Heat Flow

Consider two bodies or two phases maintained at temperatures \(T_1\) and \(T_2\), with

\[
T_2>T_1
\]

Let an infinitesimal quantity of heat \(d_iq\) be transferred from phase II at \(T_2\) to phase I at \(T_1\). Since heat is transferred from the higher temperature to the lower temperature, this is an irreversible process.

For a system undergoing an irreversible process, the entropy change is written as

\[
dS=d_eS+d_iS
\]

For heat transfer, the entropy exchanged with the surroundings is associated with the heat transfer term. For the two phases, the entropy changes resulting from the transfer of \(d_iq\) are considered separately.

The entropy gained by phase I at temperature \(T_1\) is

\[
dS_1=\frac{d_iq}{T_1}
\]

Phase II loses the same amount of heat. Therefore, its entropy change is

\[
dS_2=-\frac{d_iq}{T_2}
\]

The total entropy change resulting from the heat transfer is therefore

\[
d_iS=dS_1+dS_2
\]

Substituting the two entropy changes,

\[
d_iS=
\frac{d_iq}{T_1}
–
\frac{d_iq}{T_2}
\]

Taking \(d_iq\) common,

\[
\boxed{
d_iS=d_iq
\left(
\frac{1}{T_1}-\frac{1}{T_2}
\right)
}
\]

Since \(T_2>T_1\),

\[
\frac{1}{T_1}>\frac{1}{T_2}
\]

and hence

\[
d_iS>0
\]

This is consistent with the second law of thermodynamics: entropy production during an irreversible process is positive.

Rate of Entropy Production

Dividing the above equation by \(dt\),

\[
\frac{d_iS}{dt}
=
\frac{d_iq}{dt}
\left(
\frac{1}{T_1}-\frac{1}{T_2}
\right)
\]

Therefore, the rate of entropy production is

\[
\boxed{
\sigma=
\frac{d_iq}{dt}
\left(
\frac{1}{T_1}-\frac{1}{T_2}
\right)
}
\]

where \(\sigma\) denotes the rate of entropy production.

This expression can be written in the general form

\[
\boxed{
\sigma=J_qX_q
}
\]

where

\[
J_q=\frac{d_iq}{dt}
\]

is the heat-flow flux and

\[
X_q=
\left(
\frac{1}{T_1}-\frac{1}{T_2}
\right)
\]

is the corresponding thermodynamic driving force.

Thus, the rate of entropy production is the product of a flux and its corresponding driving force.

Condition for Zero Entropy Production

From

\[
d_iS=d_iq
\left(
\frac{1}{T_1}-\frac{1}{T_2}
\right)
\]

entropy production becomes zero when

\[
\frac{1}{T_1}-\frac{1}{T_2}=0
\]

Therefore,

\[
\frac{1}{T_1}=\frac{1}{T_2}
\]

and hence

\[
\boxed{T_1=T_2}
\]

Thus, entropy production due to heat flow becomes zero only when the two phases reach thermal equilibrium.

At thermal equilibrium, there is no temperature difference and consequently no net heat flow caused by that temperature difference.

The result can therefore be summarized by

\[
\boxed{
\text{Temperature difference}
\rightarrow
\text{heat flow}
\rightarrow
\text{entropy production}
}
\]

and at equilibrium,

\[
\boxed{
T_1=T_2
\quad\Rightarrow\quad
J_q=0
\quad\Rightarrow\quad
d_iS=0
}
\]