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Homework Help: What is the change in internal energy of the water

  1. Jan 18, 2010 #1
    Hi there, I have some serious problems with a couple of questions. Need help for exam revision.
    1.1 The problem statement, all variables and given/known data
    A 2kg chunck of ice at -10Celsius is added to 5kg of water at 45Celsius. What is the final temperature of the system?

    1.2 Relevant equations
    specific heat capacity of ice - 2100 Jkg-1K-1
    specific heat capacity of water - 4190 Jkg-1K-1
    latent heat of fusion for ice - 334 000 Jkg-1


    1.3 The attempt at a solution
    Using formula E=mC(change in)T
    E1 - 2 x 2100 x 10 = 42000J
    E2 - 2 x 334 000 = 668 000J
    E3 - 668 000 + 42 000 = 710 000J
    E4 - 710 000=5 x 4190 x (change in)T T=33.89Celsius, so final temperature is 45-33.89=11.11, but the correct answer has to be 7.9Celsius.

    2.1 The problem statement, all variables and given/known data
    A cylinder with a piston contains 0.2 kg of water at 100 Celsius. What is the change in internal energy of the water when it is converted to steam at 100Celsius at a constant pressure of 105 Nm-2?

    2.2 Relevant equations
    latent heat of vaporisation of water - 2.26 x 106 Jkg-1
    density of water - 1000 kgm-3
    density of steam - 0.6 kgm-3

    2.3 The attempt at a solution
    Using (change in) U=(change in)Q - P x (change in)V
    U=internal energy, Q=heat, P=pressure, V=volume
    Change in heat is 2.26 x 106 x 0.2 = 452 000 J
    Can't manage more :(.
    Supposed answer is 418.7 kJ.
     
  2. jcsd
  3. Jan 19, 2010 #2

    ehild

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    Re: Thermodynamics

    "E4 - 710 000=5 x 4190 x (change in)T"

    This is not correct.

    You have to warm up to the 2kg 0 oC melted ice to the final temperature.
     
  4. Jan 19, 2010 #3
    Re: Thermodynamics

    Sorry, I can't see it still. I know it has taken 710 000Joules of energy to get the ice to 0 Celisu and melt completely to water. So I thought i simply take that energy away from the water at 45 Celsius. Thus using Energy=mass*specific heat capacity of water* change in temperature to get the change in temperature from 45.
     
  5. Jan 19, 2010 #4

    ehild

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    Re: Thermodynamics

    It is true that you need that 710000 J to warm up and melt the ice to oC water. That energy is taken away from the 45 oC water, but you still have 2 kg 0 oC water to be warm up to the final temperature.

    ehild
     
  6. Jan 19, 2010 #5
    Re: Thermodynamics

    Thank you Ehil, I got the answer to question 1 now.
    Worked out solution to question 2 as well.
    Change in heat is 0.2 x 2.26 x 106 Jkg-1=452 000 Joules
    Pressure is already given at 100 000 Pa
    Original volume is mass/density, so 0.2/1000=0.0002m3.
    Mass stays the same, so 0.2kg of steam occupies: 0.2/0.6=0.333333 m3 of space. Change in volume is the volume of steam - volume of water, 0.333333-0.0002=0.333133m3.
    So using the overall equation: 452 0000J - 100 000Pa x 0.333133m3=418 686.7 J.
    Can close the thread.
     
    Last edited: Jan 19, 2010
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