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

  1. Aug 13, 2014 #1
    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 = [itex]\frac{dQ}{T}[/itex]
    ΔS = [itex]\frac{Q}{T}[/itex]
    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 = [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(Tf/Ti)

    ∴ΔS1 = (0.01)(2100)ln([itex]\frac{273}{268}[/itex]) = 0.388 JK-1


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

    ΔS2 = [itex]\frac{Q}{T}[/itex] = [itex]\frac{3340}{273}[/itex] = 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. jcsd
  3. Aug 13, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. Aug 14, 2014 #3
    Great, thanks
     
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