I don't know how I got "mysterious energy" in a Carnot heat pump

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SUMMARY

The discussion centers on the operation of a Carnot heat pump, specifically addressing how the heat delivered to the hot reservoir (Q_H) can exceed the work input (W). The equations derived include |Q_H| = |W| * (K + 1), where K represents the efficiency coefficient, and |Q_L| = K * |W|. The participant confirms that the heat pumped into the hot reservoir is a combination of the work done and the heat extracted from the cold reservoir, leading to a net gain in energy. This understanding clarifies the apparent paradox of energy output exceeding input in a Carnot cycle.

PREREQUISITES
  • Carnot refrigeration cycle principles
  • Understanding of thermodynamic efficiency coefficients
  • Basic knowledge of heat transfer concepts
  • Familiarity with thermodynamic equations and transformations
NEXT STEPS
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  • Learn about the implications of the second law of thermodynamics on heat pumps
  • Explore real-world applications of Carnot heat pumps in HVAC systems
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in the design or analysis of heat pump systems will benefit from this discussion.

Philip Robotic
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Homework Statement
Calculate the heat Q_H that a Carnot heat pump can deliver to the reservoir with a temperature T_H. Temperature of lower reservoir (T_L) and work (W, W>0) are given
Relevant Equations
Equations come from Carnot refrigeration cycle
1. |W|=|Q_H|-|Q_L|
2. K=|Q_L|/|W|
So first I transformed the equation no 2 like this:
$$|Q_L|=K\cdot|W|$$
And then I transformed the first equation to find ##|Q_Z|##
$$|Q_L|=|Q_H|-|W|$$
Plugging the result into the first equation
$$|Q_H|=K\cdot |W|+|W|$$
$$|Q_H|=|W|\cdot (K+1)$$
We know that the efficiency coefficient K is greater than 0, so how is it possible that the energy "pumped" into the hot reservoir (##Q_H##) is greater than the work that was put into the system?
 
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Philip Robotic said:
Homework Statement: Calculate the heat Q_H that a Carnot heat pump can deliver to the reservoir with a temperature T_H. Temperature of lower reservoir (T_L) and work (W, W>0) are given
Homework Equations: Equations come from Carnot refrigeration cycle
1. |W|=|Q_H|-|Q_L|
2. K=|Q_L|/|W|

So first I transformed the equation no 2 like this:
$$|Q_L|=K\cdot|W|$$
And then I transformed the first equation to find ##|Q_Z|##
$$|Q_L|=|Q_H|-|W|$$
Plugging the result into the first equation
$$|Q_H|=K\cdot |W|+|W|$$
$$|Q_H|=|W|\cdot (K+1)$$
We know that the efficiency coefficient K is greater than 0, so how is it possible that the energy "pumped" into the hot reservoir (##Q_H##) is greater than the work that was put into the system?
The heat to the hot reservoir is the work that was put into the system plus the heat that is pumped from the cold reservoir.
 
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So I will be able to heat up a certain substance more by using a Carnot Heat Pump than by theoretically converting all the available energy to heat? Do I understand it correctly?
 
Philip Robotic said:
So I will be able to heat up a certain substance more by using a Carnot Heat Pump than by theoretically converting all the available energy to heat? Do I understand it correctly?
There is also energy available from the cold reservoir, and you’re using some of that too.
 
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Did you use the information that it is a Carnot cycle?
 

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