Carnot Cycle efficiency, temperature dependency

[thinkingcap81]

Hi,

We know that a heat engine working on a Carnot Cycle the efficiency is:

1 - Q_L/Q_H = 1 - T_L/T_H where T is in Kelvin.

But if we use a different absolute temperature scale whose values at T_H and T_L are different, then the value of efficiency also changes.

I am confused about this issue. Where is my reasoning incorrect?

 

[russ_watters]

But if we use a different absolute temperature scale whose values at T_H and T_L are different, then the value of efficiency also changes.

Does it? Try picking some numbers and plugging them in...

 

[thinkingcap81]

Does it? Try picking some numbers and plugging them in...

Yep you're right. If we take a linear scale then the ratio $$\frac{T_L}{T_H}$$ remains constant.

 

[thinkingcap81]

A related question:

We have $$\frac{Q_H}{Q_L}=\frac{f(T_H)}{f(T_L)}$$ where T is the thermodynamic temperature. I read somewhere that Kelvin wanted a logarithmic scale where absolute 0 can be at $$-\infty$$ in the logarithmic scale. Does this mean that T must be in some linear absolute scale? If so then we cannot write that $$1-\frac{T_L}{T_H}=1-\frac{\log(T_L)}{\log(T_H)}$$ So how can i resolve this issue?

 

[vanhees71]

This cannot be right. How should it?

Another question is, why using (up to an arbitrary factor, which determines the unit of temperature, which you can as well and more conveniently measure in energy units by setting the conversion factor $k_{\text{B}}=1$ (Botlzmann constant)) the absolute temperature and not Kelvin's suggestion with a logarithm of it. The answer is of course that it's more convenient.

E.g., for a monatomic ideal gass the internal energy is given by
$$U=\frac{3}{2}N k_{\text{B}} T.$$
It would be much more complicated and of no other advantage to deal with a logarithm here.

 

[thinkingcap81]

Hi vanhees71,

Yes we could measure temperature in energy units and we would then need to calibrate such a thermodynamic thermometer against a known set of values like a monoatomic ideal gas as you said, though i am unable to visualize how to actually compare it against a real substance and then tell the temperature of the real,substance. We may not know the various contributions to the internal energy.

My query here was to understand how efficiencies of a Carnot Engine can be the same if we compare values like $$1-\frac{T_L}{T_H} \text{ and } 1-\frac{f(T_L)}{f(T_H)}$$ If as Kelvin said, absolute zero in a logarithmic scale can be at $$-\infty$$, this automatically means that the temperature used in $$f(T)$$ is the same as that used in the linear Kelvin scale.

Or am i missing something?

 

[vanhees71]

Well, if you'd have a different measure of temperature (in principle you could take any one-to-one function of the usual absolute temperature $T$), then you'd get this function into Carnot's efficiency formula. To see this, note how the efficiency is calculated, using the 1st and 2nd Law of Thermodynamics. Usually it reads
$$\mathrm{d}U=T \mathrm{d} S - p \mathrm{d} V,$$
or in any other equivalent way with the many different thermodynamical potentials all related to each other by Legendre transformations. If you'd redefine the temperature, which here is seen to be an integrating factor for heat, introducing the state quantity entropy, then you'd have to write $f(T)$ instead of $T$. As I said you don't gain anything with that but rather make things more complicated.

 

[thinkingcap81]

Well, if you'd have a different measure of temperature (in principle you could take any one-to-one function of the usual absolute temperature $T$), then you'd get this function into Carnot's efficiency formula. To see this, note how the efficiency is calculated, using the 1st and 2nd Law of Thermodynamics. Usually it reads
$$\mathrm{d}U=T \mathrm{d} S - p \mathrm{d} V,$$
or in any other equivalent way with the many different thermodynamical potentials all related to each other by Legendre transformations. If you'd redefine the temperature, which here is seen to be an integrating factor for heat, introducing the state quantity entropy, then you'd have to write $f(T)$ instead of $T$. As I said you don't gain anything with that but rather make things more complicated.

We could be making things more complicated. I just wanted to know what Kelvin had in mind when he proposed using the logarithmic scale with absolute 0 at -∞