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468 CHAPTER 20 IRREVERSIBLE THERMODYNAMICS
System 2 Conducting rod System 1
J Q
System boundary Cold
for rod
Hot T 2 T 1
l
Temperature T
Distance
FIGURE 20.1
Steady-state conduction of heat along a bar.
Consider the system shown in Fig. 20.1, in which a thermally insulated rod connects two reservoirs
at temperatures T 2 and T 1 respectively. Heat flows between the two reservoirs by conduction along the
rod. If the reservoirs are very large the conduction of energy out of, or into, them can be considered not
to affect them. A state will be achieved when the rate of heat flow, dQ/dt, entering the rod equals the
rate of heat flow, dQ/dt, leaving the rod. If a thermometer was inserted at any point in the rod the
reading would not change with time, but it would be dependent on position. If the rod was of uniform
cross-section, the temperature gradient would be linear. (This is the basis of the Searle’s bar conduction
experiment.)
The temperature in the bar is therefore a function of position but is independent of time. The overall
system is in a ‘stationary’ or ‘steady’ state but not in ‘equilibrium’ for that requires that the temperature
be uniform throughout the system. If the metal bar was isolated from all the influences of the sur-
roundings and from the heat sources, i.e. if it is made an isolated system the difference between the
steady state and equilibrium becomes obvious. In the case where the system was in steady state,
processes would occur after the isolation (equalisation of temperature throughout the bar); where the
system was already in equilibrium, they would not.
20.2 ENTROPY FLOW AND ENTROPY PRODUCTION
Still considering the conduction example given above. If the heat flows into the left-hand end of the bar
due to an infinitesimal temperature difference, i.e. the process is reversible, the left-hand reservoir
loses entropy at the rate
dS 2 1 dQ
¼ (20.1)
dt T 2 dt
Similarly the right-hand reservoir gains entropy at the rate
dS 1 1 dQ
¼ (20.2)
dt T 1 dt

