(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)?
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 :)