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Latent Heat and Phase Changes

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    25 kg of ice at 0C is combined with 4 kg of steam at 100C. What is the final equilibrium temperature in Celsius of the system? (Lf = 3.33x 10^5 J/kg; Lv= 2.26x10^6 J/kg; 1 cal= 4.186 J) Answer: 20


    2. Relevant equations
    mc(Tf- Ti)
    m(Lf) or m (Li)


    3. The attempt at a solution
    I thought you should do an equation for both parts so:
    m= mass c = specific heat T= temperature
    mc(Tf-Ti) + m(lf) +mc(Tf-i)=0
    25(2090)(Tf-0) + 25(3.33x10^6) + 25(4186)(Tf-0) = 0

    then...
    mc(Tf-Ti) + m(Lv) +mc(Tf-i)=0
    4(2010)(Tf-100)+ 4(2.26x 10^6) + 4(4186)(Tf-100) = 0

    however I used these equation and solved for Tf in both and added them together to get a final answer and the answer was not 20.

    please help!! thanks in advanced.
     
  2. jcsd
  3. Jun 22, 2008 #2

    Hootenanny

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    Why should these equations be equal to zero? What do these equations represent?
     
  4. Jun 22, 2008 #3
    The first equation is the the ice changing to water
    and the second is the steam changing to water in the system.
    The equation equal zero because you are finding equilibrium. (???)
     
  5. Jun 22, 2008 #4

    Hootenanny

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    Yes but what quantity do these equations represent, what are you adding up?
     
  6. Jun 22, 2008 #5
    i think its total amount of energy in Joules (I'm not sure what you are looking for)
     
  7. Jun 22, 2008 #6

    Hootenanny

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    Correct! But you need to be careful with your signs. Let's build up the equations one at a time starting with that of ice. What are the energy changes that will occur to ice?
     
  8. Jun 22, 2008 #7
    25(2090)(Tf-0) + 25(3.33x10^6) + 25(4186)(Tf-0) = 0


    wait...is it
    25(2090)(Tf-0) - 25(3.33x10^6) + 25(4186)(Tf-0) = 0

    ??not sure what is going on
     
  9. Jun 22, 2008 #8

    Hootenanny

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    Let's start from the top: you have a block of ice at 0oC and you mix it with steam at 100oC. What happens to the ice (in terms of phase transitions and energy changes)?
     
  10. Jun 22, 2008 #9
    Oh yeah it does say this under the problem... hint: assume both ice and steam become water.
    so would I find mc(delta t) +mL + mc(delta T) for ice and subtract that from the steam
     
  11. Jun 22, 2008 #10

    Hootenanny

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    We're heading roughly in the right direction. Could you explain to me what each of those terms represents [i.e. mc(delta t) +mL + mc(delta T)]?
     
  12. Jun 22, 2008 #11
    Ok ...well first for the ice

    mc(delta t)
    the transferred energy from freezing to ice. so you would use 25 kg for ice, c (specific heat) = 2090 for ice. and the initial temperature is 0 and final is unknown.

    Then you have to take in account the fact that when the ice is 0 degrees mixture remains at this temperature until all the ice melts so using the equation mLf (Heat of fusion) is 25(3.33 x 10^5 J/kg)

    except this just made me realize the ice is at 0 degrees (so do I not have to do that first part??)

    and then I was going to do mc(delta T) for the ice to turn to water
    25(4186)(Tf - inital) would initial still be 0, right?


    I was going to do the same process with the steam except 4kg adn heat of vaporization

    mc(delta t) +mL + mc(delta T)
     
  13. Jun 22, 2008 #12
    ok I think i figured out the answer:

    25(3.33x10^5) +25(4186)Tf -4(2.26x10^6) + 4(2.26x10^6) + 4(4186)(Tf-100) =0

    I solve for Tf and get 19.6 which is approx. 20.

    except I'm not sure I fully understand it...
    could you explain why you do not do 4(2010)(tf -100) (when the water is at boiling.)
    is it because you are just doing what changes so you do not have to do the current state.
    also could you explain why you subtract 4(2.26x10^6)


    thank you so much for your help!!
     
  14. Jun 23, 2008 #13

    Hootenanny

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    This isn't correct. Let's step back a bit to your previous post.
    That is correct. There is no need to calculate the energy associated with the change in temperature of the ice since there is no temperature change (the ice must change state first). What you are calculating is the energy change of the water, or the energy required to melt the ice and then raise the temperature of the resulting water to a given temperature Tf. So explicitly, the energy used/gained by the ice/water is:

    [tex]\Delta Q_\text{ice} = mL_\text{ice} + mc_\text{water}\left(T_f - 0\right)[/tex]

    [tex]\Delta Q_\text{ice} = mL_\text{ice} + mc_\text{water}T_f[/tex]

    Do you follow? Can you now do the same for the steam?
     
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