Question about the Carnot engine

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?

Boltzmann2012

I'm not much of a specialist, but I think the fact can be deduced easily while one works through the derivation of carnot efficiency.

Here is the link from wikipedia.

http://en.wikipedia.org/wiki/Carnot_...d_significance

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 T_H and T_C. It was, I believe, the idea of William Thomson (Lord Kelvin) to define temperature such that [tex]\frac{T_H}{T_C}=\frac{Q_H}{Q_C}[/tex] in which Q_H and Q_C 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.

Andrew Mason

Start with:

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

titaniumpen

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