Question

A reversible heat engine, operating in a cycle, withdraws thermal energy from a high-temperature reservoir (the temperature of which consequently decreases), performs work w, and rejects thermal energy into a low-temperature reservoir (the temperature of which consequently increases). The two reservoirs are, initially, at the temperatures T1 and T2 and have constant heat capacities C1 and C2, respectively. Calculate the final temperature of the system and the maximum amount of work which can be obtained from the engine.

Answer 1

The final temperature is
`\left(T_1^(C_1)+T_2^(C_2)\right)^{\frac {1}{C_1+C_2}}`

and the maximum amount of workdone is
`C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^(C_1)+T_2^(C_2)\right)^{\frac {1}{C_1+C_2}}`.

Step-by-step explanation:

Assume that
`R_1` is the reservior having temperature
`T_1 K` and heat capicity
`C_1 \frac JK` and
`R_2` is the reservior having temperature
`T_2` and heat capicity
`C_2 \frac JK`.

The work will be extracted till that both the reservior reach the thermal equilibrium. Let the final temperature of both the reservior is
`T_f`.

Let total
`Q_1` heat is extracted by the heat engine from the reservior
`R_1` and its temperature decreases from
`T_1` to
`T_f` and
`Q_2` heat is rejected by the heat engine to the reservior
`R_2` and its temperature decreases from
`T_2` to
`T_f`.

So, The maximum amount of work done,

`w= Q_1 - Q_2\; \cdots (i)`

Now, as the heat engine is reversible, so change is entropy for the universe is 0, which means sum of change in entropy for the ststem as well as surrounding is 0.

As shown in figure, the system is the reversible engine, so, change is entropy for the system is 0. Hence, change in entropy for the the surrounding is 0.

As temperature of
`R_1` is changing fron
`T_1` to
`T_f`, so, change in entropy of surrounding due to transfer of
`Q_1` is
`C_1 \ln \frac {T_f}{T_1}`.

Similarly, change in entropy of surrounding due to transfer of
`Q_2` is
`C_2 \ln \frac {T_f}{T_2}`.

As the net change in entropy of the surrounding is 0.

`\Rightarrow C_1 \ln \frac {T_f}{T_1}+C_2 \ln \frac {T_f}{T_2}=0`

`\Rightarrow \ln \left( \frac {T_f}{T_1} \right)^(C_1)+ \ln \left( \frac {T_f}{T_2}\right)^(C_2)=0`

`\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^(C_1)=- \ln \left( \frac {T_f}{T_2}\right)^(C_2)`

`\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^(C_1)= \ln \left( \frac {T_2}{T_f}\right)^(C_2)`

`\Rightarrow \left( \frac {T_f}{T_1}\right)^(C_1)=\left( \frac {T_2}{T_f}\right)^(C_2)`[taking anti-log both the sides]

`\Rightarrow T_f^((C_1 +C_2))=T_1^(C_1)+T_2^(C_2)`

`\Rightarrow T_f=\left(T_1^(C_1)+T_2^(C_2)\right)^{\frac {1}{C_1+C_2}}\; \cdots (ii)`

This is the required final temperature.

Now, from equarion (i), the maximum amount of work done is

`w= Q_1 - Q_2`

As
`Q=C\Delta T`

`\Rightarrow w=C_1(T_1-T_f)-C_2(T_f-T_2)`

`\Rightarrow w=C_1T_1 + C_2T_2-(C_1+C_2)T_f`

From equation
`(ii)`,

`w=C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^(C_1)+T_2^(C_2)\right)^{\frac {1}{C_1+C_2}}`

This is the required maximum workdone.