# Thermodynamics -- Temperature of a Heat Source?

Summary:
2nd law of thermodynamics and Carnot heat engine
In heat engine we define a heat source from where heat is transferred to the system, we say that heat source has a temperature ##T_h## , When we define a Carnot heat engine, the first process we have is an isothermal expansion and we say heat has to come in system through this process and here the ##T_h## is the constant temperature of that isothermal process .Are they both same? I'm getting confused in getting what actually is the source and sink in both cases and what are their temperature? Also in the entropy equation $$\oint dQ_r/T=0$$ what is the temperature T?

• 256bits

Chestermiller
Mentor
The temperature of the heat source and the system temperature are both the same value in the isothermal heating segment of the Carnot cycle. In the "entropy equation" that you have written, T is the hot reservoir temperature (and system temperature) during the isothermal heating (expansion) segment, and T is the cold reservoir temperature (and system temperature) during the isothermal cooling (compression) segment.

• • Rahulx084, 256bits, berkeman and 1 other person
That mean heat source and system temperature are two different cases, but the Carnot heat engine is a special case where both reservoir and system temperature are same? Can you give me one practical example sir to elaborate this source ,sink ,system and their temperature concept.

Chestermiller
Mentor
That mean heat source and system temperature are two different cases, but the Carnot heat engine is a special case where both reservoir and system temperature are same? Can you give me one practical example sir to elaborate this source ,sink ,system and their temperature concept.
Please describe for me in your own words the details of what is happening in each of the 4 segments of the Carnot cycle.

In the first segment : Reversible Isothermal expansion
3rd:Reversible Isothermal compression

Chestermiller
Mentor
How do the constant gas temperatures in steps 1 and 3 compare? What is the direction of heat transfer in step 1, and what is the direction of heat transfer in step 3? How do the amounts of heat transferred in steps 1 and 3 compare?

I don't get what is meant by constant gas temperature comparison of step 1 and 3 . Rest of it is as follows:
Heat is getting into the system in step 1
Heat is getting out of system in step 2
We can find the heat transfer in step 1 and 3 using the work done formula, as in reversible isothermal process heat transfer = work done
$$Q=W=nRTln(V_2/V_1)$$

Chestermiller
Mentor
In step 1, the temperature of the hot reservoir (and the gas) are higher than in step 3 where the temperature of the cold reservoir (and the gas) are lower. Actually, in step 1, the temperature of the hot reservoir is slightly higher than the average gas temperature and, in step 3, the temperature of the cold reservoir is slightly lower than the average gas temperature; that way, heat can be transferred. Is this what you were wondering about.

So for the cycle, $$\Delta S_{gas}=\frac{Q_{hot}}{T_{hot}}-\frac{Q_{cold}}{T_{cold}}=0$$ Is this what you were asking?

We know that during a phase change operations ##\Delta S##= ##m\lambda / T_{sat}##
Where ##dQ=m\lambda##
How we have determined that T here? Like whats the source here or the sink ?

Chestermiller
Mentor
We know that during a phase change operations ##\Delta S##= ##m\lambda / T_{sat}##
Where ##dQ=m\lambda##
How we have determined that T here? Like whats the source here or the sink ?
You are aware that, in the standard Carnot cycle, there is no change of phase, right?

For a reversible expansion involving a working fluid phase change, to be reversible, the hot reservoir temperature must be slightly higher than the saturation temperature of the working fluid at the pressure of the expansion. Is that what you are asking?

We know that during a phase change operations ##\Delta S##= ##m\lambda / T_{sat}##
Where ##dQ=m\lambda##
How we have determined that T here? Like whats the source here or the sink ?
I means lets not consider any carnot cycle here , just for a reversible phase change process .

Chestermiller
Mentor
I means lets not consider any carnot cycle here , just for a reversible phase change process .
For the change to be reversible, the temperature of the source must be slightly higher than the saturation temperature of the working fluid that is changing phase; otherwise, if the hot reservoir temperature is higher than this, there will be temperature gradients within the working fluid (giving rise to entropy generation), and $$\Delta S_{fluid}=\frac{m\lambda}{T_{sat}}>\frac{m\lambda}{T_h}$$ where ##Q=m\lambda##. So, for the irreversible case, the change in entropy of the working fluid will be greater than the heat transferred from the hot reservoir divided by the hot reservoir temperature. In the application of the Clausius inequality, you always use the temperature at the boundary between the system and the surroundings to divide the heat transferred at the boundary. In the present case, this is the same as the hot reservoir temperature.

Thank you so much sir , I get it now , one final doubt , I'm attaching a picture, I don't want to know the solution of it , just tell me that what will be the ##Q_h## here , like the practical meaning of it , like ##Q_c## is the heat removed from refrigerator.

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Chestermiller
Mentor
Thank you so much sir , I get it now , one final doubt , I'm attaching a picture, I don't want to know the solution of it , just tell me that what will be the ##Q_h## here , like the practical meaning of it , like ##Q_c## is the heat removed from refrigerator.
##Q_h## is the heat rejected to the room, and is equal to $$Q_h=Q_c+W=5\ kW + W$$

If you look under the refrigerator, you will see a set of tubular coils over which room air is blown by a fan to remove the heat. The heat transfer occurs at the room air temperature.

Correct me If I'm wrong ,
Lets say a refrigerator is working at 5 degree Celsius and we need it at 3 degree Celsius, so heat required to be removed to bring 5 to 3 degree Celsius is the ##Q_c## and for to remove this heat we have to do some work thats W so ##Q_c## + W is the ##Q_h## and this ##Q_h## is the one what we feel warm outside a refrigerator .

Chestermiller
Mentor
Correct me If I'm wrong ,
Lets say a refrigerator is working at 5 degree Celsius and we need it at 3 degree Celsius, so heat required to be removed to bring 5 to 3 degree Celsius is the ##Q_c## and for to remove this heat we have to do some work thats W so ##Q_c## + W is the ##Q_h## and this ##Q_h## is the one what we feel warm outside a refrigerator .
If you wanted to lower the temperature inside the refrigerator from 5 C to 3 C, you would have to increase ##Q_c## above 5 kw, and this would require a proportional increase in both W and in ##Q_h##. Regarding ##Q_h##, it is what you feel when you put your hand in the air stream coming from the outside coils.

Okay sir , thank you very much again :)