# Obtaining work from two bodies by a heat engine

• Parzeevahl
In summary, two bodies are heated by a heat engine, with net work obtained as the sum of the individual works. The efficiency of the engine is given by 1 - (T_f/T_initial), but this equation is not applicable in this context. The maximum efficiency corresponds to no entropy gain, and the assumption that all heat transfer takes place at T1 and T2 is incorrect. The maximum efficiency is achieved when there is no entropy change for the combination of hot and cold bodies.
Parzeevahl
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. 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?

Not sure where you are getting that equation for the "most efficient " condition from, but I would say it does not apply in this context.
Instead, use that maximum efficiency corresponds to no entropy gain.

You are going wrong is implicitly assuming in that all the heat transfer takes place at T1 and T2. That's what your two efficiency equations assume. But, in this case, heat transfer from the hot body takes place at all temperatures between T1 and Tf, and heat transfer to the colder body takes place at all temperatures between T2 and Tf. Haruspex is correct in saying that the maximum efficiency corresponds to no entropy change for the combination of hot and cold bodies.

## 1. How does a heat engine obtain work from two bodies?

A heat engine obtains work from two bodies by utilizing the difference in temperature between the two bodies. It takes in heat energy from a high temperature reservoir and releases some of that energy to a lower temperature reservoir, thus converting heat energy into mechanical work.

## 2. What is the principle behind obtaining work from two bodies by a heat engine?

The principle behind obtaining work from two bodies by a heat engine is the Second Law of Thermodynamics. This law states that heat will naturally flow from a higher temperature to a lower temperature, and a heat engine takes advantage of this natural flow to convert heat energy into work.

## 3. Can a heat engine obtain work from two bodies indefinitely?

No, a heat engine cannot obtain work from two bodies indefinitely. The Second Law of Thermodynamics also states that in any energy conversion, some energy will be lost as heat. This means that a heat engine will eventually reach a state of equilibrium where it can no longer produce work.

## 4. What are the components of a heat engine that allow it to obtain work from two bodies?

A heat engine typically consists of a heat source, a heat sink, and a working fluid. The heat source is where the high temperature reservoir is located, while the heat sink is where the lower temperature reservoir is located. The working fluid is a substance that can be heated and cooled to convert heat energy into mechanical work.

## 5. Can a heat engine obtain work from two bodies at any temperature difference?

No, a heat engine can only obtain work from two bodies if there is a temperature difference between them. The greater the temperature difference, the more work can be obtained. However, there is a limit to how much work can be obtained, and this limit is determined by the efficiency of the heat engine.

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