14)Part 1: Five thousand kilocalories (5000 kcal) of heat flows from a high temperature reservoir at 600 K to a low temperature reservoir at 300 K.What is the entropy change in the universe?Part 2: Now let us assume the same 5000 kcal of heat flows through a 20% efficient heat engine between the same reservoirs.a. Compute amount of heat converted into useful work.b. Compute the entropy change in the universe. Equations(average speed) d= vot + }at² (distance under acceleration)FnetW= mg (Weight) a=(Acceleration) F = Gmm2(Gravity Law)(Мотentum) КЕ%3D3mu?(Kinetic Energy) PE =mgh (Potential Energy)p= mvW = F ·d (Work Done) Q= mcAT (Heat Energy) Q=mL (Latent Heat Energy)eff = (Heat Engine Efficiency) eff = 1- (2nd Heat Engine Efficiency)W%3DeffCarnot = 1 – (Efficiency Carnot Cycle) S = $ (Entropy Change of Reservoir)AS = H + c (Total Entropy Change Closed System)

Entropy change is related to the amount of heat transferred at a temperature T as, ∆S=∆QT∆Q is the amount of heat transferred at temperature T For any reservoir at a given temperature, heat can be absorbed or released by this reservoir without any changes in its temperature. The amount of heat transferred or released from the hot reservoir is 5000 kcal ∆Q=5000 kcal∆Q=5000×4184=2.092×107 JThe change in entropy for the hot reservoir is ∆S=-∆QT∆S=-2.092×107600∆S1=-34866.67 J/K The negative sign is due to the fact that heat is being released from this reservoir. This heat released by the hot reservoir is absorbed by the cool reservoir, at temperature 300 K. So the entropy change for this reservoir is ∆S2=2.092×107300∆S2=69733.33 J/K Here, a positive sign indicates that heat is absorbed. The total change in the entropy of the universe is given as ∆S=∆S1+∆S2∆S=-34866.67+69733.33∆S=34866.66 J/KNow, between the same two reservoirs is a heat engine, with a 20% efficiency. This heat engine operates between the temperatures 600 K and 300 K So the heat engine takes 5000 kcal of heat from the reservoir at 600 K, does some work and then discards the rest of the remaining heat to the lower temperature 300 K reservoir. In doing so, the efficiency of the heat engine is given as η=WQ1W is the amount of work done by the engineQ1 is the amount of heat absorbed by the engine The heat absorbed by the engine is the heat taken from the hot reservoir, 5000 kcal Thus, Q1=5000 kcal=2.092×107 J And efficiency, η=20%=20100=0.2Thus, the amount of work done by this heat engine is η=WQ1W=ηQ1W=0.2×2.092×107W=4184000 J And in doing this work, the rest of the remaining heat is discarded to the lower temperature reservoir, which is given as Q2=Q1-WQ2=2.092×107-4184000Q2==1.6736×107 J This is the amount of heat absorbed by the lower temperature reservoir Thus, now the entropy changes of the high and low temperature reservoirs respectively are ∆S1=-2.092×107600∆S1=-34866.67 J/K∆S2=1.6736×107300∆S2= 55786.667 J/Kwe know that as the carnot cycle is a perfectly reversible process, its entropy change is zerotherefore∆Se = 0 And so the entropy change of the universe is ∆S=∆S1+∆S2+ ∆Se∆S=-34866.67+55786.667 + 0∆S=20920 J/KAnswers: Part 1: Entropy change of the universe is 34866.66 J/K Part 2: Work done by the heat engine is 4184000 J Entropy change of the universe is 20920 J/K

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Advanced Physics

Part 1: Five thousand kilocalories (5000 kcal) of heat flows from a high temperature reservoir at 600 K to a low temperature reservoir at 300 K. What is the entropy change in the universe? Part 2: Now let us assume the same 5000 kcal of heat flows through a 20% efficient heat engine between the same reservoirs. a. Compute amount of heat converted into useful work. b. Compute the entropy change in the universe.

Part 1: Five thousand kilocalories (5000 kcal) of heat flows from a high temperature reservoir at 600 K to a low temperature reservoir at 300 K. What is the entropy change in the universe? Part 2: Now let us assume the same 5000 kcal of heat flows through a 20% efficient heat engine between the same reservoirs. a. Compute amount of heat converted into useful work. b. Compute the entropy change in the universe.

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Question

14)

Part 1: Five thousand kilocalories (5000 kcal) of heat flows from a high temperature reservoir at 600 K to a low temperature reservoir at 300 K.

What is the entropy change in the universe?

Part 2: Now let us assume the same 5000 kcal of heat flows through a 20% efficient heat engine between the same reservoirs.

a. Compute amount of heat converted into useful work.

b. Compute the entropy change in the universe.

Equations
(average speed) d= vot + }at² (distance under acceleration)
Fnet
W
= mg (Weight) a=
(Acceleration) F = Gmm2
(Gravity Law)
(Мотentum) КЕ%3D3mu?
(Kinetic Energy) PE =mgh (Potential Energy)
p= mv
W = F ·d (Work Done) Q= mcAT (Heat Energy) Q=mL (Latent Heat Energy)
eff = (Heat Engine Efficiency) eff = 1- (2nd Heat Engine Efficiency)
W
%3D
effCarnot = 1 – (Efficiency Carnot Cycle) S = $ (Entropy Change of Reservoir)
AS = H + c (Total Entropy Change Closed System)

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