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

[physics.stu]

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.

 

[rude man]

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.

 

[physics.stu]

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)

 

[DrClaude]

Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir

So what is the efficiency?

 

[rude man]

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!