Carnot Refrigerator Work Input: 230 J Extracts 346 J Heat

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Discussion Overview

The discussion revolves around the operation of a Carnot refrigerator, specifically addressing the relationship between work input and heat extraction. Participants explore the implications of energy conservation in the context of thermodynamic systems, questioning how a refrigerator can extract more heat than the work input it receives.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question whether the scenario presented implies a violation of the law of conservation of energy.
  • Others argue that the first law of thermodynamics is not violated because the heat rejected (Qh) is greater than the heat extracted (Qc).
  • A participant expresses confusion about how 230 J of work can result in 346 J of heat, suggesting that according to conservation of energy, the work should convert directly to an equivalent amount of heat.
  • One participant explains that the work done is necessary to move heat against the temperature gradient, indicating that the gas transitions from 346 J to 576 J with the input of work.
  • Another participant uses an analogy involving lifting weight to illustrate the concept of work done in relation to energy extracted.
  • There is a mention of the need for a formula to calculate work done in a different context, suggesting a broader exploration of energy transfer concepts.
  • A later reply emphasizes that moving energy within a system does not necessarily require more energy than the energy being moved, challenging assumptions about energy costs in thermodynamic processes.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the Carnot refrigerator's operation, with no consensus reached on whether the scenario violates conservation laws or how the energy transfer mechanisms function.

Contextual Notes

Some participants reference the need for mathematical models and equations related to Carnot engines, indicating that there may be unresolved assumptions or steps in the calculations discussed.

Mohammed Alqadhi
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If a Carnot refrigerator requires a work input of only 230 J to extract 346 J of heat from the cold reservoir.
Doesn’t this discrepancy imply a violation of the law of conservation
of energy?
 
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Does it? Please state your reasoning.
 
My reasoning to that is it does not violate the 1st law because the heat Qh that will be rejected will be bigger than the extracted Qc.
what I wondering is how 230J of work results in 346J of heat?
According to the conservation of energy, 230J of work will be converted,If so, to 230J of heat or any other kind of energy, is that right?
 
Mohammed Alqadhi said:
My reasoning to that is it does not violate the 1st law because the heat Qh that will be rejected will be bigger than the extracted Qc.
Why not look up the carnot engine equations and do some math?
what I wondering is how 230J of work results in 346J of heat?
According to the conservation of energy, 230J of work will be converted,If so, to 230J of heat or any other kind of energy, is that right?
The work is the extra energy that has to come from outside in order to get the heat to flow "backwards" against the temperature gradient between the two reservoirs ... the gas starts out with 346J, and ends up with 576J. You had to do 230J of work to do that. The simple model does not tell you exactly how this happened.
 
Last edited:
I may not deliver my question right or you may not got it.
Anyway I did the math and it get out right, but I am asking that the work applied to the refrigerator to extract the heat from the cold staff should not be equal or more than the heat extracted?
 
It's like carrying a 5kg weight up 4.7m of stairs ... that would mean you have to do 230.3J of work.
If the bottom of the stairs was 7.3m above sea level, maybe the house is on a hill, then the energy of that 5kg weight was 357.7J
So you just did about 230J of work to extract 357.7J from the bottom of the stairs.
 
I understand this, but how is that related?
 
How much gasoline does it take to transport 1000 gallons of gasoline 100 yards down the road?
 
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We get the volume and the distance, if I am correct about it what's the formula that gives us the work done, so we can find the amount of gasoline by using Q=mLc?
??
 
  • #10
I don't want you to get bogged down in the detail. The point I am making is that that moving energy from one part of a system to another (eg from cold to hot reservoir) is not the same as adding energy to a system. There is no law that says moving energy costs more energy than the energy being moved.

A lorry might consume less than 1gallon moving 1000gallons that short distance.

Perhaps worth reading up on "closed systems" as conservation of energy only applies to those. If you don't include a hot reservoir in your system it's not closed.
 
  • #11
This helps a little, but I am still confused!
 

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