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

[Pouyan]

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?)

 

[TSny]

Isn't heat being added to the ice rather than "taken from the ice"?

Let dQ_H and dQ_C 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 dQ_C in terms of η and dQ_H.

 

[Pouyan]

Isn't heat being added to the ice rather than "taken from the ice"?

Let dQ_H and dQ_C 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 dQ_C in terms of η and dQ_H.

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?

 

[Chestermiller]

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?

 

[Pouyan]

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