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A heat engine operates between a hot reservoir with a temperature of 400 K and a cold reservoir with a temperature of 300 K. The engine accepts 500 J of energy from the hot reservoir, does 150 J of work, and exhausts 350 J of energy into the cold reservoir.1)Does this engine violate any laws of thermodynamics?It violates neither the first nor the second law of thermodynamics.It violates the first law of thermodynamics, but not the second.It violates the second law of thermodynamics, but not the first.It violates both the first and second laws of thermodynamics. Can someone explain?

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Answer:

It violates the second law of thermodynamics, but not the first.

Explanation:

Given:

temperature of hotter reservoir, [tex]T_H=400\ K[/tex]

temperature of colder reservoir, [tex]T_L=300\ K[/tex]

Energy consumed by the engine, [tex]Q_H=500\ J[/tex]

heat rejected by the engine, [tex]Q_L=350\ J[/tex]

work output of the engine, [tex]W=150\ J[/tex]

  • The First law of thermodynamics is in synchronization with the law of conservation of energy and for this cycle:

Total energy input = Total energy output

[tex]Q_H=W+Q_L[/tex]

[tex]500=150+350[/tex] which is true.

  • Heat engine is device operating in a continuous cycle between two reservoirs such that it transfers heat from a high temperature reservoir to a low temperature reservoir giving some work output in synchronization to Kelvin-Plank statement of the second law of thermodynamics.

Now we compare the actual efficiency and the ideal (Carnot) efficiency of this cycle.

Actual efficiency:

[tex]\eta_a=\frac{W}{Q_H}[/tex]

[tex]\eta_a=\frac{150}{500}[/tex]

[tex]\eta_a=0.3\ or\ 30\%[/tex]

Ideal efficiency:

[tex]\eta_c=1-\frac{T_L}{T_H}[/tex]

[tex]\eta_c=1-\frac{300}{400}[/tex]

[tex]\eta_c=0.25\ or\ 25\%[/tex]

Here we observe that ([tex]\eta_c<\eta_a[/tex]) ideal efficiency is less than the actual efficiency which violates the Second Law of thermodynamics.

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