Solving a 0.5kg Copper Heat Transfer Problem

[CatWoman]

Homework Statement

(a) A 0.5kg mass of copper (specific heat 385 J kg-1K-1) at 600K is plunged into a litre of water at 20C. What is the equilibrium temperature, T2, of the system? What is the change in entropy of the water?
(b) A litre of water is heated slowly at a constant rate from 20C to T2. What is the change in entropy?
How do you account for the difference between (a) and (b)?

Homework Equations

not totally sure

The Attempt at a Solution

(a) ∆T_water=T_2-T_1=T_2-293 so ∆Q_water=C ∆T=〖4200 (T〗_2-293)

∆T_copper=T_2-T_1=T_2-600 so ∆Q_copper=C ∆T=〖385/2 (T〗_2-600)

Heat lost by copper is heat gained by water so ∆Q_water= ∆Q_copper so

〖4200 (T〗_2-293)=〖385/2 (T〗_2-600) => T_2=336K=63 degrees Centigrade

This is not an isothermal process so cannot use S=Q/T to calculate entropy before and after to calculate entropy difference.
Instead, use ∆S=C ln⁡〖T_2/T_1 〗 so ∆S=4200 ln⁡〖336/293=575 J K^(-1) 〗
However, if I treat it like an isothermal process, where
∆Q=4200 (336-293)=4200×43=180600 Joules
Then ∆S=∆Q/T=180600/293=616 J K^(-1)
The extra entropy gained could be because some of the water has turned to steam.

Please could somebody tell me what I am doing right, how I should calculate the entropy differently for part a and part b, and why, as I am very confused!

Many thanks :)

 

[Mapes]

Entropy is a state variable, dependent only pressure, temperature, and amount of material. I'm having a hard time seeing why the entropy change of water between 20°C and some arbitrary temperature $T_2$ could be path dependent, even if some quantity has boiled and recondensed.

 

[CatWoman]

Many thanks - yuor logic makes sense to me so I don't know what the question is trying to get at.