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Thermodynamics - ice->water->steam

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A 8.0 g ice cube at -30 C is in a rigid, sealed container from which all the air has been evacuated.

    How much heat is required to change this ice cube into steam at 240 C?

    2. Relevant equations

    Q_net = 0 where Q is heat

    C1 = Specific heat capacity, ice: 2.050 kJ/kg-K = 2050 J
    C2 = Specific heat capacity, water: 4.184 kJ/kg-K = 4184 J
    C3 = Specific heat capacity, water vapor: 1.996 kJ/-kgK = 1996 J

    L_f = Latent heat fusion, water 334*10^3 J/kg
    L_v = Latent heat vaporization, water 2257*10^3 J/kg

    Q_net = Q1 + Q2 + Q3 = 0
    Q1 = -30deg C ice --> 0deg C ice, Q2 = 0deg C water --> 100deg C water, Q3 = 100deg C vapor --> 240deg C vapor

    Q_net = (C1)(M)(T1) + M(L_f) + (C2)(M)(T2) + M(L_v) + (C3)(M)(T3) = 0
    M = 6.0 grams = 0.006 kg
    T1 = 30 deg
    T2 = 100 deg
    T3 = 140 deg



    3. The attempt at a solution

    Q_net = (C1)(M)(T1) + M(L_f) + (C2)(M)(T2) + M(L_v) + (C3)(M)(T3)

    = (2050)(0.006)(30) + (0.006)(334*10^3) + (4184)(0.006)(100) + (0.006)(2257*10^3) + (1996)(0.006)(140)

    = 369 + 2004 + 2510.4 + 13542 + 1995.16

    = 20420.56 Joules

    did i account correctly for all the phase changes etc? is my final answer correct? any help appreciated.

    thanks
     
  2. jcsd
  3. Sep 8, 2008 #2

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    It looks correct except m = 0.008kg, not 0.006kg.
     
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