- #1

Parzeevahl

- 6

- 0

- Homework Statement
- Two identical bodies of constant heat capacity Cp at temperatures T1 and T2 respectively are used as reservoirs for a heat engine. If the bodies remain at constant pressure, show that the amount of work obtainable is

W = Cp * (T1 + T2 - 2 * Tf)

where Tf is the final temperature attained by both bodies. Show that if the most efficient engine is used, then Tf^2 = T1 * T2.

- Relevant Equations
- W = Cp * dT

Here's my attempt for the first part:

For the first body, the work obtained is

##W_1 = C_P (T_1 - T_f)##

while for the second body, it is

##W_2 = C_P(T_2 - T_f).##

So the net work obtained is the sum of these two:

##W = W_1 + W_2 = C_P (T_1 + T_2 - 2 T_f)##

and that proves the first part.

##1 - \frac{T_{final}}{T_{initial}}.##

There are two heatings here, for which we find the efficiencies to be

##1 - \frac{T_f}{T_1}##

and

##1 - \frac{T_f}{T_2}.##

But the heat engine is the same so they should be the same, which means

##T_1 = T_2##

but that was not the case originally! Where have I gone wrong?

For the first body, the work obtained is

##W_1 = C_P (T_1 - T_f)##

while for the second body, it is

##W_2 = C_P(T_2 - T_f).##

So the net work obtained is the sum of these two:

##W = W_1 + W_2 = C_P (T_1 + T_2 - 2 T_f)##

and that proves the first part.

**However, here's my problem with the second part:**The same heat engine is doing all the job,*i.e.,*heating the two bodies. Now, the question asks me to consider the**most efficient engine**, in which case the efficiency is##1 - \frac{T_{final}}{T_{initial}}.##

There are two heatings here, for which we find the efficiencies to be

##1 - \frac{T_f}{T_1}##

and

##1 - \frac{T_f}{T_2}.##

But the heat engine is the same so they should be the same, which means

##T_1 = T_2##

but that was not the case originally! Where have I gone wrong?