# Is efficiency of Carnot engine dependent on temperature of heat source/sink?

Suppose we are working with a Carnot engine. In this engine, temperature of the heat source is say $$373K$$ and that of heat sink is say $$293K$$. Now, the initial temperature of the gas(ideal) is 333K. The gas is isothermally and adiabatically expanded, and its temperature was reduced to 313K, and again the gas is compressed adiabatically and isothermally raising its temperature to 333K. So, a reversible cycle is completed. Now, if we calculate the efficiency of the engine then we will obtain, efficiency=1-Q2/Q1.

Here, Q1→heat absorbed in isothermal expansion step

Q2→heat rejected in isothermal compression step.

We know that in case of ideal engine/carnot engine heat is proportional to temperature which is in our case Q1∞333K and Q2∞313K as per the decription in the link below:

https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node24.html

But as per the image attached below this shows that Q1∞373K(temperature of the heat source) and Q2∞293K(temperature of the heat sink).

So my question is when the cylinder comes in contact with the heat source is it let to increase its temperature to that of the temp. Of the source or it can be any temperature? If it can be any then why does most of the books and articles state that efficiency of the carnot engine is dependent on tbe temperature of heat source and heat sink? • "Suppose we are working with a Carnot engine. In this engine, temperature of the heat source is say - $373K$ and that of heat sink is say - $293K$." I assume these are hyphens and not minus signs since you can't get below 0K Apr 10, 2021 at 16:07
• Yes these are hyphens... Didn't notice that when edit was suggested.
– MSKB
Apr 10, 2021 at 16:09
• In order for the cycle to be reversible, the temperature of the gas has to always be infinitesimally less than the temperature of the hot reservoir and infinitesimally more than the temperature of the cold reservoir. So the gas temperature cannot be initially 333K. Apr 10, 2021 at 16:11
• Is this phenomenon maintained to proceed isothermally?
– MSKB
Apr 10, 2021 at 16:17
• See my answer and the update responding to your last comment. Apr 10, 2021 at 16:20

If the Carnot engine is hotter than the high-temperature reservoir, then when they are brought into thermal contact, heat will not flow from the reservoir to the engine. It will flow in the other direction: from the engine to the reservoir. So you will not get any work output from such a Carnot engine (but you might provide work input and use it as a heat pump).

When we use the Carnot cycle to provide work output, we want the heat flow at both reservoirs to be reversible (in the thermodynamic sense of the word). The heat flow will only be reversible if the temperature difference between reservoir and engine is infinitesimal while they are in thermal contact.

If the heat flow is not reversible then two things follow. First, the standard derivation of the Carnot theorem will show that the efficiency cannot exceed that of a reversible process. Secondly, the entropy of the engine will go up more than the minimum for the given amount of heat it receives, so it will have to get rid of more heat to the lower reservoir in order to restore its entropy to the starting value. Hence either observation can be used to show that such a process is less efficient than a reversible process.

So my question is when the cylinder comes in contact with the heat source is it let to increase its temperature to that of the temp. Of the source or it can be any temperature?

To reiterate my comment, in order for the cycle to be reversible, the temperature of the gas has to always be infinitesimally less than the temperature of the hot reservoir and infinitesimally more than the temperature of the cold reservoir. Any finite temperature difference between the gas and the reservoirs results in irreversible heat transfer.

So if the gas temperature is initially 333K and the hot reservoir 373K, the heat addition is irreversible.

Is this phenomenon maintained to proceed isothermally?

Yes, it is required for the process to be a reversible isothermal process. The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic (isentropic) processes.

Hope this helps.

• Ohh I get it. The more is the temperatuee difference the more distant equilibrium is. Since isothermal process is carried out very slowly a series of short distance equilibrium is to be maintained otherwise this process cannot be isothermal and that is why the temperature difference should be minimum. Is this notion correct?
– MSKB
Apr 10, 2021 at 16:20
• @MohammadSakibShahriar Exactly, and well put. Apr 10, 2021 at 16:21