How Do You Determine Temperature in Series Carnot Engines by Equating Work Done?

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Prabs3257
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Homework Statement
Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two engines A and B are represented by ηA and ηB, respectively,then what is the value of Ratio of ηA and ηB
Relevant Equations
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I know i have to use the efficiency formula and everything is fine but i don't know how to find T its the only unknown in my equation can someone please tell me how to find T . In the solution they got the value of T by equating the work done by the two engines , but why is their work done equal ?
 
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There is nothing in your statement of the problem that constrains T. Either the answer is supposed to leave T as an unknown parameter, or there must be more to the question.
 
In terms of T, if QH is the heat received by engine A at 600 K, what is the amount of heat Q rejected to engine B? In terms of the amount of heat Q received by engine B at T, what is the amount of heat rejected by engine B at 100 K?
 
Assuming equating the work of the two engines? If T is close to 600K the hotter engine will obviously do less work than the cooler. Etc. Therefore, not a good assumption!

But - I suspect this was postulated as part of the problem, in which case T can indeed be found in terms of T1 and T2 (600K and 100K).

Hint: if each engine is Carnot, what can you conclude from considering the series-connected engines as one engine?
 
Hi Prabs3257. The key here is to realize that the heat flow into the reservoir from engine A is exactly equal to the heat flow out of the reservoir to engine B. So there is no accumulation of heat in that middle reservoir at temperature T. This means that the total entropy change is 0. Therefore, the two engines operate as equivalent to a single carnot engine operating between 600K and 100K.

AM
 
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rude man said:
Assuming equating the work of the two engines?
I suspect this was postulated as part of the problem
Apparently so! Found this on the net

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