WOLFRAM|DEMONSTRATIONS PROJECT

Heat Transfer and the Second Law of Thermodynamics

​
Δq J
100
T
1
K
500
T
2
K
300
ΔS =
Δ
q
1
​
T
1
+
Δ
q
2
​
T
2
= 0.133 J/K
Two thermal reservoirs at the same temperature,
T
2
=
T
1
, can, in concept, exchange an increment of heat
Δq
in either direction. Since it can just as easily flow in the opposite direction, this represents what is known as a reversible process and
Δq
can be designated
Δ
q
rev
. By convention,
Δq
is positive if heat flows into a reservoir and negative (
-Δq
) if it flows out. When
T
2
>
T
1
, it is a matter of experience that
Δq
flows spontaneously (or irreversibly) from the hotter to the cooler reservoir. One can still set
Δq=Δ
q
rev
if one conceptualized an infinite number of intermediate reservoirs at temperature increments varying infinitesimally between
T
2
and
T
1
.
An increment
Δq
represents the process of heat exchange between two systems. Neither system can be said to possess a quantity of heat
q
. In the usual terminology,
q
is not a "function of state" for a thermodynamic system. However,
d
q
rev
/T
turns out to represent the differential of a new function of state,
dS
, where
S
is known as the entropy. A derivation of this relation is given in Details for the case of an ideal gas.
A compact statement of the second law of thermodynamics is that the entropy of an isolated system (or of the whole universe) always increases in a spontaneous process,
ΔS>0
. The reverse process is unattainable—as Bertrand Russell put it, "You cannot unscramble eggs." The limiting case
ΔS=0
pertains to reversible processes (themselves an idealization).
An operational statement of the second law is due to Clausius: "It is impossible in a cyclic process to produce no other effect than the transfer of heat from a cooler to a warmer body." Of course, refrigerators do this all the time, but this requires coupling to other devices, such as motors and condensers.
In this Demonstration, you can vary the temperatures
T
2
and
T
1
and the quantity of heat transferred,
Δq
. The total entropy change of the system is then calculated using
ΔS=
Δ
q
1
T
1
+
Δ
q
2
T
2
. Recall that
Δq
is counted as negative when a reservoir gives off heat. The system of two reservoirs is presumed to be isolated from its surroundings.