# Entropy change of melting ice cube initially at -5°C

1. Aug 13, 2014

### Flucky

1. The problem statement, all variables and given/known data
Calculate the entropy change of an ice cube of mass 10g, at an initial temperature of -5°C, when it completely melts.

cice = 2.1 kJkg-1K-1
Lice-water = 3.34x105 Jkg-1

2. Relevant equations
dQ = mcdT
dS = $\frac{dQ}{T}$
ΔS = $\frac{Q}{T}$
Q = mL

3. The attempt at a solution
First I set the problem out in two stages:
a) the entropy change from the ice going from -5°C to 0°C (in order to melt)
b) the entropy change from the ice going to water

For a)
dQ = mcdT ---------(1)
dS = $\frac{dQ}{T}$ ---------(2)

Putting (1) into (2):

dS = $\frac{mcdT}{T}$
ΔS = mc∫$\frac{1}{T}$dT
ΔS = mcln(Tf/Ti)

∴ΔS1 = (0.01)(2100)ln($\frac{273}{268}$) = 0.388 JK-1

For b)
Q = mL = (0.01)(3.34x105) = 3340J

ΔS2 = $\frac{Q}{T}$ = $\frac{3340}{273}$ = 12.23 JK-1

∴ total ΔS = ΔS1 + ΔS2 = 0.388 + 12.23 = 12.62 JK-1

Am I right in simply adding the to changes of entropy together? Does ΔS work like that?

Cheers.

2. Aug 13, 2014

### rude man

Looks good, and the answer is yes, the entropies add. Entropy is a state function, like gravitational potential. If you went from 0 to 1m above ground you would have g x 1m change in potential. If you went from 1m to 2m there would be a further g x 1m change in potential. Giving total change in potential = g x 2m.

3. Aug 14, 2014

### Flucky

Great, thanks