Thermodynamics and entropy of reservoirs

In summary, the amount of work obtainable from two identical bodies at constant heat capacity and temperatures T1 and T2 is given by W = cp(T1 + T2 - 2Tf)(T2 - T1), where Tf is the final temperature attained by both bodies and cp is the heat capacity at constant pressure. This is obtained by using the equation for heat flow into a reservoir and considering the net heat equal to the net work for a cycle.
  • #1
2DGamer
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1. Two identical bodies at constant heat capacity and temperatures T1 and T2 are used as reservoirs for a heat engine. If the bodies remain at constant pressure and undergo no change of phase, show that the amount of work obtainable is: W = cp(T1+T2-2Tf) where Tf is the final temperature attained by both bodies and cp is the heat capacity at constant pressure.



2. cp = heat capacity at constant pressure
ΔS = -Q/Tf = -cp(Tf - Ti)/Tf (entropy change in reservoir)




3. I'm having a bit of trouble with this problem. I figured I'd start by making a T-S diagram using the T1 and T2 temperatures where T2 > T1. Then the net heat (Q) would be ΔS(T2-T1). Since ΔU = 0 for a cycle the net heat would have to equal the net work. So, W = ΔS(T2-T1). To find ΔS I used the equation for a heat flow into a reservoir which is ΔS = -cp(Tf - Ti)/Tf. So for both reservoirs I have ΔS1 = -cp(Tf - T1)/Tf and ΔS2 = -cp(Tf - T2)/Tf. Then I said: ΔS = ΔS1 + ΔS2. I find this to be: ΔS = (cp/Tf)(T1 + T2 -2Tf). Which looks similar to what I'm supposed to be getting. But it can't be right because I still have to multiply that by (T2 - T1) to get the work obtainable from the above equation. Can someone please help me out with this? What am I doing wrong? Thanks!
 
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  • #2
The equation you derived is correct. To get the work obtainable, you just need to multiply it by (T2 - T1). So, the amount of work obtainable is W = cp(T1 + T2 - 2Tf)(T2 - T1).
 

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships between heat, energy, and work. It studies the behavior of systems that involve changes in temperature, pressure, and volume.

2. What is a reservoir in thermodynamics?

In thermodynamics, a reservoir is a large system that can supply or absorb an infinite amount of heat without changing its temperature. It serves as a reference point for measuring the thermodynamic properties of other systems.

3. How does entropy relate to thermodynamics?

Entropy is a measure of the disorder or randomness of a system. In thermodynamics, it is often used to describe the direction and efficiency of energy transfer and the availability of energy to do work.

4. What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system always increases over time. This means that systems tend to become more disordered and less able to do work without an input of energy.

5. How do reservoirs affect the entropy of a system?

Reservoirs play a crucial role in maintaining the second law of thermodynamics. They act as sources or sinks of energy, allowing systems to reach equilibrium and increasing the overall entropy of the universe.

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