Proving a heat engine more efficient than Carnot violates the second law

[jlmac2001]

I'm having trouble with proving the following question. Can someone please help, please.

(1) Prove that if you had a heat engine whose efficiency was better than the ideal value (4.5) you could hook it up to an ordinary Carnot refrigerator to make a refrigeratior that requires no work input.


I know that efficency = 1-Qc/Qh and COP=1/(Qh/Qc - 1)but how do i prove the above question?

 

[turin]

Originally posted by jlmac2001
... if you had a heat engine whose efficiency was better than the ideal value (4.5) ...

Either I'm really ignorant of this thermo stuff, or you've gotten the wrong number for efficiency. I think it should always be less than unity.

Originally posted by jlmac2001
... you could hook it up to an ordinary Carnot refrigerator ...

My first step would be to draw a diagram. Make sure to include the hot and cold reservoirs and the directions of energy/heat flow. That should give a good picture of what's going on.

 

[M98Ranger]

To prove this question, we can use the concept of the second law of thermodynamics, which states that no heat engine can have an efficiency greater than that of a Carnot engine operating between the same temperature limits. This means that for a heat engine with an efficiency greater than the ideal value of 4.5, its efficiency must be greater than the efficiency of a Carnot engine operating between the same temperature limits.

Let's assume that the heat engine in question has an efficiency of x, which is greater than 4.5, and let's also assume that the Carnot engine has an efficiency of y. According to the second law of thermodynamics, we know that x > y.

Now, if we hook up the heat engine with efficiency x to an ordinary Carnot refrigerator, the heat rejected by the heat engine (Qc) would be equal to the heat absorbed by the refrigerator (Qh). This is because the heat engine is operating at a higher efficiency, so it would reject less heat than the Carnot engine.

Using the equations you mentioned, we can write the efficiency of the refrigerator as y = 1 - Qc/Qh. Substituting Qc = Qh, we get y = 1 - 1 = 0. This means that the efficiency of the refrigerator is 0, which implies that no work input is required for the refrigerator to operate.

Therefore, by hooking up the heat engine with efficiency x to an ordinary Carnot refrigerator, we have created a refrigerator that requires no work input, thus proving the statement.