# Homework Help: Total entropy change for ice melting in a room

1. Aug 12, 2012

### plpg

1. The problem statement, all variables and given/known data
Okay, so I am having difficulties with understanding the concepts around entropy, take this question:
What is the total entropy change for 7 kg of ice melting from -5 C° to 5 C° in room at 5 C°.

2. Relevant equations
dS=dQ/T
Q =mΔH
m*c*ln(tfinal/tinitial)
c_ice=2
c_water=4
c_air=1

3. The attempt at a solution
Okay, so this is my reasoning so far:
as the ice heats up to zero centigrades, the total change in entropy is given by integrating dS=dQ/T wrt T, and we end up at : m*c*ln(tfinal/tinitial), so 7*2*ln(273.15/(273.15-5)) j/k.

As the ice changes its state to water, the change in entropy is given directly by dS=dQ/T, where dQ is =mdH=7*330=2310 j/k.

When the water heats up from 0 centigrades to 5 centigrades its change in entropy is again given by, 7*4*ln(273.15+5/(273.15)) j/k,
and adding these we get the total entropy change for the melting ice.

Now to the part that I am having big difficulties understanding: the change in entropy of the room. I've read that it is given by m*c_air*((tfinal-tinitial)/tfinal), so in this case this would be negative 7*1*((273.15+5°-(273.15-5°))/(273.15+5°)) j/k.

So adding all these would give me the answer? 7*2*ln(273.15/(273.15-5)) j/k + 2310 j/k + 7*4*ln(273.15+5/(273.15)) j/k - 7*1*((273.15+5°-(273.15-5°))/(273.15+5°)) j/k.

Am i even close to getting this question right?

Last edited by a moderator: Aug 13, 2012
2. Aug 12, 2012

### Andrew Mason

This is correct except for the units. What are the units for the specific heat of ice?
Again, correct except for the units.
Units.

You have to assume that the temperature of the room air does not change. The room is large enough that the heat flow from the air changes the temperature of the air by a negligible amount. So the entropy change is just$\Delta S_{room} = \Delta Q_{room}/T_{room}$ where $\Delta Q_{room} = - \Delta Q_{ice}$.
AM

3. Aug 12, 2012

### plpg

1234

Last edited: Aug 12, 2012
4. Aug 12, 2012

### Andrew Mason

If you are using Kg and a specific heat of 2 KJ/Kg for ice and 4 KJ/Kg for water (which I assume is the case since you are using 7 and not 7,000 as the mass), the units for entropy will be in KJ/K.

Correct so far.

If you are using mass in Kg, the specific heat is in units of KJ/Kg. For example, the total heat flow required to melt 7kg of ice is 7*330 KJ = 2310 KJ = 2,310,000 J.

AM

5. Aug 13, 2012

### Staff: Mentor

Cheating by deleting your posts after you have received help is strictly against the PF rules. Check your PMs, and do not do this again here.