AZING! Understanding the Carnot Cycle: Heat, Work, and Internal Energy

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SUMMARY

The discussion focuses on the Carnot Cycle, specifically the transitions between points in the cycle concerning heat (Q), internal energy (U), and work (W). During the higher temperature isothermal phase, heat enters the system, resulting in positive work done by the system with no change in internal energy. Conversely, in the cooler isothermal phase, heat exits the system, necessitating work done on the system. In the adiabatic phases, the relationship between work and internal energy is defined by the equation ΔU = -W, indicating no heat flow.

PREREQUISITES
  • Understanding of thermodynamic concepts, particularly the Carnot Cycle.
  • Familiarity with the first law of thermodynamics.
  • Knowledge of isothermal and adiabatic processes.
  • Basic grasp of heat transfer mechanisms.
NEXT STEPS
  • Study the first law of thermodynamics in detail.
  • Learn about isothermal and adiabatic processes in thermodynamics.
  • Explore practical applications of the Carnot Cycle in heat engines.
  • Investigate the efficiency calculations of Carnot engines.
USEFUL FOR

Students of thermodynamics, engineers working with heat engines, and anyone seeking to deepen their understanding of the Carnot Cycle and its implications in energy systems.

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Homework Statement


I don't understand the process between the points in a carnot cycle.


Homework Equations


Can someone please explain what is going on from point to point in terms of Q (heat), U(thermal energy), and Work?


The Attempt at a Solution

 
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Wikipedia's page on the http://en.wikipedia.org/wiki/Carnot_cycle" does a good job of explaining what is occurring in terms of heat flow and work.

In the higher temperature isothermal part, heat flows into the system. Since it is isothermal, there is no change in U so \Delta Q = W (expansion-positive work done by the system). In the cooler isothermal part heat flows out of the system with no change in U so work must be done on the system (compression). Of course in the adiabatic parts there is no heat flow so the work done + the change in internal energy = 0: \Delta U = -W.

AM
 
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