Thermal Question -- Carnot Engine

[StrSpeed]

Homework Statement

A Carnot engine extracts 745 J from a 592-K reservoir during each cycle and rejects 485J to a cool reservoir. It operates 16 cycles per second. Find work, efficient, temp of cool reservoir.

Homework Equations

W = Qh - Qc
n=Wout/Qh

The Attempt at a Solution

I'm sure I can do the problem I just don't understand what the problem means by rejects. If someone could explain possibly that would be absolutely fantastic. I've spent a good deal of time reading and googling but I am still missing it.

I just can't seem to figure out how to get the 485J and 592K to relate.

Thank you so much!

I've gotten now :
W = 745J - 485J | This I know is right

Qc = Qh - W ===> Qc = 592K-W (from above)
I'm not sure this is right, and the answer isn't in the back of the book.

 

[Luis Obis]

Let my try to clarify it for you.

A heat engine is a machine that transforms heat to work. You should know (2º law of thermodynamics) that heat cannot be completely transformed to work, there has to be some "wasted heat" in this case the heat flowing to the cold reservoir. Otherwise you would be violating the second law. You can prove that maximum efficiency is achieved if the process is reversible, like on this case, a Carnot cycle, and you can also derive the expression for the efficiency of the carnot cycle, which is always less than 1, like it's supposed to.

$\eta = \frac{W}{Q_H} = \frac{Q_H-Q_C}{Q_H} = 1 - \frac{T_C}{T_H} < 1$

You can see a complete derivation of this expresion here:

https://en.wikipedia.org/wiki/Carnot_heat_engine

As you can see, from this formula you can get the efficiency from the work which you know how to calculate, then you can use this to get the temperature of the cold reservior.

I hope this helped you!

 

[haruspex]

I just don't understand what the problem means by rejects.

It just means the heat that flows to the cold reservoir.

 

[StrSpeed]

Thank you two both so much. I feel ridiculous now.