A firebox is at 650 K and the ambient temperature is 250 K T

A firebox is at 650 K, and the ambient temperature is 250 K. The efficiency of a Carnot engine doing 146 J of work as it transports energy between these constant-temperature baths is 61.5%. The Carnot engine must take in energy

146 J/0.62 = 237.3 J from the hot reservoir and must put out 91.3 J of energy by heat into the environment. To follow Carnot\'s reasoning, suppose some other heat engine S could have efficiency 70.0%.(a) Find the energy input and exhaust energy output of engine S as it does 146 J of work.

(b) Let engine S operate as in part (a) and run the Carnot engine in reverse between the same reservoirs. The output work of engine S is the input work for the Carnot refrigerator. Find the total energy transferred to or from the firebox and the total energy transferred to or from the environment as both engines operate together.

(c) Explain how the results of parts (a) and (b) show that the Clausius statement of the second law of thermodynamics is violated.


(d) Find the energy input and work output of engine S as it puts out exhaust energy of 91.3 J.

(e) Let engine S operate as in part (d) and contribute 146 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

(h) Explain how the results show that the Kelvin—Planck statement of the second law is violated. Therefore, our assumption about the efficiency of engine S must be false.


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

(j) Explain how the result of part (i) shows that the entropy statement of the second law is violated.

Solution

(a)

Qh = W/e = W / [1 - (Tc/Th)] = 146 / [1 - (250/650)] = 237.25 J

Qc = Qh - W = 237.25 -146 = 91.25 J

Qh,net = W/e = 146/0.7 = 208.57 J

Qc,net = Qh - W = 208.57 -146 = 62.57 J

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(b)

in this case,

Qh,net = 208.57 J - 237.25 J = -28.68 J

Qc,net = 62.57 J - 91.25 J = -28.68 J

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(c)

from the above results, it is clear that the net heat flow from the cold reservoir to hot reservoir

is not possible without applying input workdone. so the second law of thermodynamics is voilated.

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(d)

Weng,S = Qc / [(1/0.7)-1] = [91.25] / [(1/0.7)-1] = 212.9 J

Qh,S = 212.9 J + 91.25 = 304.15 J

NOTE: remaining questions are posted seperately.