Thermodynamics: entropy and engines

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SUMMARY

The discussion focuses on a reversible heat engine operating between a heat source at 400 K and a heat sink at 300 K. The engine receives 1000 J of heat energy and exhausts 700 J, resulting in a work output of 300 J per cycle. The change in entropy for the heat source is calculated as 2.5 J/K, while the heat sink's change in entropy is 2.33 J/K. The engine does not violate the second law of thermodynamics, as it loses heat to perform work.

PREREQUISITES
  • Understanding of the first and second laws of thermodynamics
  • Familiarity with the concepts of heat engines and reversible processes
  • Knowledge of entropy calculations and units (J/K)
  • Basic algebra for solving equations related to thermodynamic cycles
NEXT STEPS
  • Study the Carnot cycle and its efficiency calculations
  • Learn about real-world applications of heat engines and their limitations
  • Explore advanced entropy concepts in thermodynamics
  • Investigate the implications of the second law of thermodynamics in various systems
USEFUL FOR

Students studying thermodynamics, engineers designing heat engines, and anyone interested in the principles of energy conversion and entropy.

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Entropy and Engines

Homework Statement


A heat engine has the following diagram associated with it. There is a heat source which supplies 1000 J of heat energy during one cycle and remains at a constant temperature of 400 K. A heat sink exhausts 700 J of heat during a cycle at a constant temperature of 300 K. Both are reversible processes. During one cycle the engine itself doesn't change its entropy.

A) How much work can this engine do in one cycle?
B) Calculate change of entropy of heat source through one cycle?
C) Calculate change of entropy of heat sink through one cycle of engine?
D) Does this engine violate second laws of thermodynamics?

Homework Equations



Q[tex]_{H}[/tex] = Q[tex]_{L}[/tex] + W
[tex]\Delta[/tex]S = Q/T

The Attempt at a Solution



I just need someone to check my work. Thanks!

A) 1000 = 700 + W
W = 300 J

B) 1000 J/ 400 k = 3 J/K (well, 2.5, but with one sigfig...)

C) 700 J/ 300 K = 2 J/K (again, 2 and 1/3, but round down...)

D) No, because heat is lost in order to do work.
 
Last edited:
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Anyone? Please? It's a simple problem...
 
And are the sigfigs right?

sorry...*blush*
 

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