- #1

Flucky

- 95

- 1

## 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

^{-1}K

^{-1}

L

_{ice-water}= 3.34x10

^{5}Jkg

^{-1}

## Homework Equations

dQ = mcdT

dS = [itex]\frac{dQ}{T}[/itex]

ΔS = [itex]\frac{Q}{T}[/itex]

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 = [itex]\frac{dQ}{T}[/itex] ---------(2)

Putting (1) into (2):

dS = [itex]\frac{mcdT}{T}[/itex]

ΔS = mc∫[itex]\frac{1}{T}[/itex]dT

ΔS = mcln(T

_{f}/T

_{i})

∴ΔS

_{1}= (0.01)(2100)ln([itex]\frac{273}{268}[/itex]) = 0.388 JK

^{-1}

__For b)__Q = mL = (0.01)(3.34x10

^{5}) = 3340J

ΔS

_{2}= [itex]\frac{Q}{T}[/itex] = [itex]\frac{3340}{273}[/itex] = 12.23 JK

^{-1}

∴ total ΔS = ΔS

_{1}+ ΔS

_{2}= 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.