What is the change in entropy of the water in a reversible heat engine?

AI Thread Summary
The discussion revolves around calculating the change in entropy of water in a reversible heat engine. The initial approach to determine heat transfer was incorrect, leading to confusion about the efficiency and heat exchange processes. The correct formula for heat transfer involves integrating the efficiency over temperature, where the negative sign indicates heat is being removed from the system. Participants clarified that in a reversible process, the change in entropy of the water and ice is equal and opposite, leading to a net change of zero. Ultimately, the correct understanding of the relationship between heat transfer and entropy change was achieved, confirming the calculations.
Pouyan
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Homework Statement
1 kg of liquid water with a temperature of 373 K is cooled in a reversible heat engine by a very large block of ice with a temperature of 273 K until the water (still liquid) adopts the temperature of 273 K. Determine how much ice is melted in the process.
Relevant Equations
C_w = C_water= 4200 (J/KgK)
dQ_water= 1kg*C_w*dT
C_ice=334000 J/kg
Q_ice= m_ice * C_ice
My attempt:
I though :
ΔQ_w= 1*4200 * (-100) J=-420000J
Q_ice=334000*m_ice = ΔQ_w

But it was totaly wrong!
The solution showed :

Because the heat engine is reversible the efficiency η = 1- (T_cold / T)
T_cold is always 273 K while the hot temperature changes from 373 K to 273 K during this loop.
The amount of heat taken from the ice :
Q = -C ∫(1-(1-(T_cold/T)))*dT [from 373 to 273]
Q=358 kJ

Q_ice=334000*mm=Q/Q_ice = 1.07 kg

My question is this part:
Q = -C ∫(1-(1-(T_cold/T)))*dT
What do I see it is Q=-C*∫(1-η)dT
1-But what does 1-η mean?
2-Where does the negative sign come from? (I mean why do we write -C?)
 
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Isn't heat being added to the ice rather than "taken from the ice"?

Let dQH and dQC denote small amounts of heat taken from the hot reservoir and given to the cold reservoir, respectively, during a small step of the process. Use conservation of energy and the definition of efficiency η to express dQC in terms of η and dQH.
 
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TSny said:
Isn't heat being added to the ice rather than "taken from the ice"?

Let dQH and dQC denote small amounts of heat taken from the hot reservoir and given to the cold reservoir, respectively, during a small step of the process. Use conservation of energy and the definition of efficiency η to express dQC in terms of η and dQH.
I tried with this but I don't know if I am right!
1- (Qc/Qh)= 1-(Tc/T)
Qc/Qh = Tc/T
Qc=Qh*Tc/T

where Qh=∫m*C*dT
Qc in this case will be -358 kJ
But I can not still understand why should we multiply it by -1?
 
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Chestermiller said:
If the engine operates reversibly, the change in entropy of the water plus the ice is zero. What is the change in entropy of the water?

Bingo!

I should think on that as well!

Now I can write in this way :

S_ice=- S_water

Where S_water= ∫m*c/T *dT!

S_ice = Q_ice/T_c !

Now I have a good solution ! Thank you for this help:smile:
 
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