(adsbygoogle = window.adsbygoogle || []).push({}); 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**