How Do You Calculate Qh for a Non-Carnot Heat Engine?

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Homework Help Overview

The discussion revolves around calculating the heat extracted from the hot reservoir (Qh) for a non-Carnot heat engine, given specific temperatures and work output. The subject area includes thermodynamics and heat engine efficiency.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between work output, thermal energy, and efficiency, questioning how to apply the first and second laws of thermodynamics. Some express uncertainty about the relevance of the Carnot engine assumptions in their calculations.

Discussion Status

Participants are actively discussing the equations related to energy conservation and efficiency, with some providing insights into the first and second laws of thermodynamics. There is a recognition that multiple approaches may be valid, and guidance has been offered to combine equations for further progress.

Contextual Notes

Some participants note the constraints of working with a non-Carnot engine and the implications of cyclic processes on energy calculations. There is also mention of the need to clarify definitions and assumptions related to efficiency.

physics.stu
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1. Thot reservoir= 600K, Tcold reservoir= 300K, Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir. What is Qh?



2. efficiency= W/Qh; Carnot efficiency= 1- (Tcold/Thot);



3. I have found the Carnot efficiency to be 0.50. I don't know where to go after this.
 
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physics.stu said:
1. Thot reservoir= 600K, Tcold reservoir= 300K, Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir. What is Qh?



2. efficiency= W/Qh; Carnot efficiency= 1- (Tcold/Thot);



3. I have found the Carnot efficiency to be 0.50. I don't know where to go after this.

1. what is the 1st law?
2. what is the 2nd law? (the assumption seems to be that you're running a Carnot engine.)
3. what is the definition of efficiency? oops, you have that equation already I see ...

3 equations, 3 unknowns.
 
First Law: dU = Q-W; dU=0.0J because it returns to initial state (cyclic engine)

I am NOT running a Carnot engine, it's the only thing that I could solve for.
Second Law: thermal energy will flow spontaneously from a hot object to cold object but cannot spontaneously flow from cold to hot object.

More on efficiency:
Qhot= W + Qcold
Efficiency= 1-(Qcold/Qhot)
 
physics.stu said:
Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir
So what is the efficiency?
 
physics.stu said:
First Law: dU = Q-W; dU=0.0J because it returns to initial state (cyclic engine)

I am NOT running a Carnot engine, it's the only thing that I could solve for.
Second Law: thermal energy will flow spontaneously from a hot object to cold object but cannot spontaneously flow from cold to hot object.

More on efficiency:
Qhot= W + Qcold
Efficiency= 1-(Qcold/Qhot)

You're right - you don't need to make the Carnot assumption. Answer Dr Claude's question!
Combine your two equations above and you have the answer!
 
Last edited:

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