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

[Flucky]

Homework Statement

Calculate the entropy change of an ice cube of mass 10g, at an initial temperature of -5°C, when it completely melts.

c_ice = 2.1 kJkg^-1K^-1
L_ice-water = 3.34x10⁵ Jkg^-1

Homework Equations

dQ = mcdT
dS = $\frac{dQ}{T}$
ΔS = $\frac{Q}{T}$
Q = mL

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(T_f/T_i)

∴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.

 

[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.

 

[Flucky]

Great, thanks