Calculating entropy change

1. Apr 2, 2012

ks23

1. The problem statement, all variables and given/known data
A 15 kg block of ice at 0 celsius degrees melts to liquid water at 0 celsius degrees inside a large room that has temperature 20 celsius degrees. Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored.
a) Is the melting of the ice reversible or irreversible?
b) Calculate the net entropy change of the system during this process.

2. Relevant equations
dS = Q/T for constant temperature
Q = m*Lf, where Lf = 3.34 x 10^5 J/kg for water
dS = int(dQ/T) for non-constant temperature

3. The attempt at a solution
I've tried to calculate the phase change from ice to water using dS=Q/T, getting 18351 J/K and then adding the entropy change for the temperature change from 0 to 20 degrees using dS=int(dQ/T)=m*c*ln(T2/T1), getting 4443 J/K. Thats no where near the correct answer, which is supposed to be 1250 J/K.

2. Apr 2, 2012

collinsmark

Hello ks23,

Welcome to Physics Forums!

So far so good.
Hold on. The problem statement doesn't say anything about letting the liquid water warm up to 20 degrees. You can stop the calculations the moment the ice completely melts. There's no need in this particular problem to let the water and room reach thermal equilibrium. The problem statement only concerns itself with the ice melting.
Don't forget about the entropy decrease of the room (ignoring the ice). The entropy of the ice==>water (ignoring the room) increases as it melts. The entropy of the room (ignoring the ice/water) decreases by some amount too because it is loosing energy that is ultimately transferred into the ice to make it water. How much energy does the room lose? (The 1st law of thermodynamics should help you with this one.) What is the temperature of the room? The overall entropy change is the sum of the ice/water's entropy change with the room's entropy change.