Carnot Cycle efficiency, temperature dependency

In summary, Kelvin's suggestion of using a logarithmic scale for absolute temperature with the value of absolute zero at -∞ would complicate things and would not provide any advantage. The efficiency of a Carnot engine can be calculated using any one-to-one function of the usual absolute temperature, but using a different scale would only make the calculations more complicated. Therefore, it is more convenient to use the traditional linear Kelvin scale for temperature measurements.
  • #1
Hi,

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

1 - QL/QH = 1 - TL/TH where T is in Kelvin.

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

I am confused about this issue. Where is my reasoning incorrect?
 
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  • #2
thinkingcap81 said:
But if we use a different absolute temperature scale whose values at TH and TL are different, then the value of efficiency also changes.
Does it? Try picking some numbers and plugging them in...
 
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  • #3
russ_watters said:
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.
 
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  • #4
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?
 
  • #5
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.
 
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  • #6
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?
 
  • #7
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.
 
  • #8
vanhees71 said:
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 -∞
 

1. What is the Carnot Cycle efficiency?

The Carnot Cycle efficiency is a theoretical maximum efficiency of a heat engine, meaning the maximum amount of work that can be obtained from a given amount of heat energy. It is expressed as the ratio of the work output to the heat input, and is denoted by the symbol η (eta).

2. How is the Carnot Cycle efficiency calculated?

The Carnot Cycle efficiency can be calculated using the formula η = 1 - (TC/TH), where TC is the temperature of the cold reservoir and TH is the temperature of the hot reservoir. This formula assumes that the Carnot Cycle is operating in a reversible and ideal manner.

3. Why is the Carnot Cycle efficiency considered the maximum efficiency?

The Carnot Cycle is considered the maximum efficiency because it is based on the fundamental principles of thermodynamics and is not affected by the specific properties of the working substance. This means that no real heat engine can have an efficiency greater than that of the Carnot Cycle.

4. How does temperature dependency affect the Carnot Cycle efficiency?

Temperature dependency plays a crucial role in the Carnot Cycle efficiency. The efficiency is directly proportional to the difference in temperature between the hot and cold reservoirs. As this temperature difference increases, the efficiency also increases, and vice versa.

5. Can the Carnot Cycle efficiency be achieved in real-life systems?

No, the Carnot Cycle efficiency cannot be achieved in real-life systems. This is because real heat engines are not completely reversible, and they also experience some energy losses due to friction and other factors. However, engineers and scientists use the Carnot Cycle efficiency as a benchmark to improve the efficiency of real heat engines.

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