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## 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 :)