A firebox is at 775 K, and the ambient temperature is 350 K. The efficiency of a

Question

A firebox is at 775 K, and the ambient temperature is 350 K. The efficiency of a Carnot engine doing 149 J of work as it transports energy between these constant-temperature baths is 54.8%. The Carnot engine must take in energy

149 J/0.55 = 271.7 J from the hot reservoir and must put out 122.7 J of energy by heat into the environment. To follow Carnot's reasoning, suppose some other heat engine S could have efficiency 70.0%.

(d) Find the energy input and work output of engine S as it puts out exhaust energy of 122.7 J.

(e) Let engine S operate as in part (d) and contribute 149 J of its work output to running the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together.
J

(f) Find the total work output.
J

(g) Find the total energy transferred to the environment.
J

(i) Let the engines operate together through one cycle as in part (d). Find the change in entropy of the Universe.
J/K

Explanation / Answer

e = W/Qh

Qh = 149/0.548

Qh = 271.9 J

d )

for engine S

|Qc ,s | = |Qh,s| -Weng,s = Wengs/es - Wengs

so work output

Weng s = |Qc,s|/( 1/e - 1) = 122.7 / [1/0.7 - 1] = 286.3 J

energy input

|Qh,s| = |Qc,s| + Weng s = 286.3 + 122.7 = 409 J

part e )

Engine S contributes 149 J out of 286.3 J to running the Carnot engine:

|Qh,net| = |Qh,s| - 271.9 = 409 - 271.9 = 137.1 J

This is the net energy lost by the firebox.

part f )

Work output

Wnet = Wengs - 149 = 286.7 - 149 = 137.7 J

part g )

|Qc,net| = 0 J

part i )

dS_total = dS_s + dS_carnot + dS_h + dS_c = 0 + 0 - 137.7/750 + 0/300

dS_total = -0.1836 J/K

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