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HW 13 Question 15

0 votes
116 views

How would one set up and approach this problem??

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.19 kJ to a cold reservoir at 27°C in each cycle.

(a) What is the engine's efficiency?

(b) How much work is done by the engine in each cycle?

(c) What is the power output of the engine if each cycle lasts 0.298 s?

asked Nov 16, 2013 in Intro College Physics by anonymous

1 Answer

0 votes
For a heat engine, we can define the heat transferred to the engine from the hot reservoir as ||H_h|| and the heat transferred from the engine to the cold reservoir as ||H_c||. The work done to the outside world by the engine is then ||W=Q_h-Q_c||. Here we all use magnitudes for work and heat.

The efficiency is calculated as $$e=1-\frac{Q_h}{Q_c}$$.

With the above relations and equations, one can get part a and b.

For part c, the definition of power is the work output per unit time ||Power=\frac {\Delta W}{\Delta t}||. In this question, the work output in a cycle is ||W=Q_h-Q_c||, divided by the time of the cycle gives the power.
answered Nov 16, 2013 by PhysicsDude (21,750 points)
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