What is the Temperature of the Low Temperature Reservoir in a Carnot Engine?

[JoeSmith1013]

Homework Statement

A Carnot engine moves a piston which applies 5N of force to a rotating disc by moving a piston 0.25cm for
a total of 5W of power. If the high temperature reservoir is 100K, what must the temperature of the low
temperature reservoir be?

Homework Equations

The efficiency of a Carnot engine is given by the formula
e = 1-TLo/THi.

The Attempt at a Solution

I know you can rearrange the efficiency equation and solve for Tlo which is TLo = (1-e)*THi. After this this step I am drawing a blank on what to do next. I know that it gives you the work done by the piston and the power of the engine, but I don't know what to do next.

 

[Simon Bridge]

Do you know anything about the carnot cycle besides the efficiency equation?

 

[rude man]

Hmm - is there a solution?
The work per carnot cycle is presumably 5N x 0.0025m = 0.0125J.
There are an infinite number of combinations of T₂ and Q₁ that can provide this work (T₂ = low temp, Q₁ = high heat) while still satisfying carnot efficiency.
The power of 5W just indicates how many carnot cycles have to be performed per second.
Comments/critiques welcome!

 

[TomHart]

Rude man/JoeSmith1013, I feel your pain on this one. To let you know where I'm at on this one, I even had to search to find out what a carnot engine was. Maybe the efficiency equation has nothing to do with it. Maybe it has to do with the fact that the gas temperature has to be reduced enough so that when it is warmed up, the gas expansion is sufficient to produce the 5N and 0.25cm. It is possible for a situation where the temperature difference is so small that the gas expansion cannot even produce the 5 N, 0.25 cm movement. So it seems to me that there has to be some minimum temperature delta to accomplish that. Just my $0.02.

 

[rude man]

Rude man/JoeSmith1013, I feel your pain on this one. To let you know where I'm at on this one, I even had to search to find out what a carnot engine was. Maybe the efficiency equation has nothing to do with it. Maybe it has to do with the fact that the gas temperature has to be reduced enough so that when it is warmed up, the gas expansion is sufficient to produce the 5N and 0.25cm. It is possible for a situation where the temperature difference is so small that the gas expansion cannot even produce the 5 N, 0.25 cm movement. So it seems to me that there has to be some minimum temperature delta to accomplish that. Just my $0.02.

There is always some minimum temperature delta required in any engine.
I did notice that the "high" temperature is pretty da**ed low (almost to frozen Nitrogen) which seems odd but I guess there it is.
Th efficiency of a carnot engine is determined by the fact that the entropy loss of the "high" reservoir = entropy gained by the low (lower!) reservoir. The possible combinations of Q₁ and T₂ I referred to previously is based on this fact.

 

[Simon Bridge]

I think there is context missing.
Its pretty specific about how the carnot cycles is being implimented so there may be an example in course notes that can help.