# Question about the Carnot engine

1. Jul 22, 2012

### titaniumpen

I was reading about the Carnot engine, and I stumbled upon this forumla:

Carnot efficiency = 1 - (QH-QL)/QH = 1 - (TH-TL)/TH

Where QH is the heat input, QL is the heat output, TH is the input temperature, TL is the output temperature.

The book says that TH is proportional to QH, and TL is proportional to QL, but it does not state why. Well it seems common sense that you have more heat input if the source is hotter, but is there a more scientific explanation? Or is it just a finding from Carnot's observations?

2. Jul 23, 2012

### Boltzmann2012

3. Jul 23, 2012

### Philip Wood

It's true by definition of thermodynamic temperature, T. You can show from the second law of thermodynamics that all Carnot engines have the same efficiency, independently of working substance, so their efficiencies can depend only on the temperatures TH and TC. It was, I believe, the idea of William Thomson (Lord Kelvin) to define temperature such that $$\frac{T_H}{T_C}=\frac{Q_H}{Q_C}$$ in which QH and QC are the heat input and heat output of a Carnot engine.

By taking the special case of an ideal gas as working substance in a Carnot engine, it is easy to show that the thermodynamic temperature as defined above, is equivalent to temperature defined by pV = nRT for a gas at limitingly low density.

Obviously what I've written is highly condensed. It is spelled out in detail in old-fashioned textbooks such as Zemansky.

4. Jul 23, 2012

### Andrew Mason

ΔS = ∫dQ/T

In the Carnot cycle, ΔS = 0. Since heat flows at constant temperature, this means that Qh/Th + Qc/Tc = 0. Since Qh = -|Qh| (i.e. heat flow is out of the hot register so it is negative), we have: |Qc|/Tc = |Qh|/Th, which reduces to |Qh/Qc|= Th/Tc.

AM

5. Jul 23, 2012

### titaniumpen

Thanks for the replies! I didn't know the answer is so simple.