Problem 10.13 (Zemansky)
Two
identical bodies of constant heat capacity at the temperatures
where
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Solution: Click For PDF Version Let the mass of each of the two bodies be m. The temperature of the body 1 is
and that of the body 2 is
 .
We assume that
 .
The two bodies are connected to an heat engine. Heat will be
extracted from body 1 and part of which will be converted into work
and part will be delivered to the body 2. This will go on till the
temperature of both bodies becomes equal, say
 .
Then the heat engine will cease to function.
Let the total
amount of heat extracted out by the heat engine from the body 1 be
As the
initial temperature of the body 1 is
Similarly the amount of heat received by the body 2 will be
Therefore, the amount of obtainable work from the two bodies will be
We next work out the change of entropy for the bodies 1 and 2. As the initial and final states of each body are well defined by their initial and final temperatures, we can compute change of entropy of each body assuming that the change takes place isobarically. We have
And
There is no change in entropy of the engine. Therefore, the total change of entropy is
From the entropy principle, we have
or
Maximum work
will be obtainable for
that is
or
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and let the amount of heat delivered by the heat engine to the body
2 be
.
The amount of work done by the heat engine in the process described
above be W. From the
energy conservation, we have
,
we have
