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Specific Heat Capacity Problem

  1. Jan 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A 30.0g block of ice at 0.00°C is dropped into 500.0g of water at 45.0°C. If the process was carried out in an iron container with a mass of 150g what would the final temperature be?

    2. Relevant equations
    Q=mc(T2-T1)
    Heat Lost = Heat Gained
    Specific Heat of Water: 4200
    Specific Heat of Ice: 2100
    Specific Heat of Iron: 460

    3. The attempt at a solution
    I understand how to find the final temperature of the water without the iron container using the equation above (mc(T2-T1)=mc(T2-T1)) and i get (34.5 degrees), however i'm completely stuck when it comes to adding the iron mass into the equation.
     
    Last edited: Jan 13, 2014
  2. jcsd
  3. Jan 14, 2014 #2
    Hi declan.B. Welcome to Physics Forums!!

    Have you learned that you are supposed to take into account the latent heat of melting the ice also? Are you supposed to assume that the starting temperature of the pot is 45C?
    Let the final temperature be T. In terms of T, how much heat do you have to add to the ice to melt it and raise its temperature to T? How much heat has to be removed from the original water in the pot and from the pot itself to cool them down from 45C to temperature T? Set these amounts of heat equal, and solve for T.

    Chet
     
  4. Jan 14, 2014 #3
    Yes sorry assuming the iron mass is also at 45 degrees initial temperature.

    I haven't found any similar questions so I'm very confused. If i was to add in the latent heat wouldn't i then need the quantity of heat which i haven't been given??

    Q= mc(t2-t1) + mLf + mc(t2-t1)

    and how would the iron mass fit into this equation ?
     
    Last edited: Jan 14, 2014
  5. Jan 14, 2014 #4

    ehild

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    The iron also cools down, and adds some heat to the ice. The ice needs heat to melt at 0°C, and then the "icewater" needs heat to warm up from zero degree to t(final).

    Meanwhile, both the water and the container cools down from 45 °C to the final temperature, releasing heat.

    Write up the equations for the heat is released both by the water and by the container.
     
  6. Jan 14, 2014 #5
    Oh gee. I guess you'll just have to look it up (in your textbook or online).
     
  7. Jan 14, 2014 #6

    SteamKing

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    And always write out the units for the quantities in your calculations.
     
  8. Jan 14, 2014 #7
    Chestermiller do you not think i have spent hours upon hours researching this. My textbook has no similar question when adding a variable like that so I have not been taught, and i cannot find it anywhere on google which is why i signed up on this forum. If you can find something teaching it please share otherwise leave your smartass comments somewhere else.

    I'm not the best at deriving equations but basically what I have gathered is what you could do is two separate equations. One to see how much heat the iron mass loses and one to see how much the water loses and then the answer is the difference? e.g.

    IRON: Q=mc(t2-t1) + mLf + mc(t2-t1)
    WATER: Q=mc(t2-t1) + mLf + mc(t2-t1)
     
  9. Jan 14, 2014 #8
    Dear declan.B,
    Please accept my apology. I resolve to avoid making sarcastic comments again in any Physics Forums. There is no place for that.

    I looked up the heat of fusion of ice to water on Google: 80 calories/gram = 334 kJ/kg.
    If T is the final temperature, then, for the part of the water that was originally ice, the amount of heat required to bring it up to the final temperature is:

    30(334)+30(4.2)(T-0) Joules

    The amount of heat that must be removed from the 150 gm iron container and the original 500 gm of water to bring their temperature from 45C to the final temperature T is:

    (150)(0.46)(45 - T) + (500)(4.2)(45-T) Joules

    The two amounts of heat must match each other.

    Note that the data provided in the problem statement regarding the heat capacity of ice is irrelevant to the solution to this problem, and was included just to distract you. You now need to solve for the temperature T.

    Chet
     
    Last edited: Jan 15, 2014
  10. Jan 14, 2014 #9

    ehild

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    Chet: check the units.

    ehild
     
  11. Jan 15, 2014 #10
    Ooops. Thanks ehild. I went back and edited in the corrections. Thanks again.

    Chet
     
  12. Jan 16, 2014 #11
    So i'm assuming you can solve both equations simultaneously as they both equal one another?
     
  13. Jan 16, 2014 #12
    Yes.

    Chet
     
  14. Jan 16, 2014 #13
    Got it. Thanks for the help :)
     
  15. Jan 17, 2014 #14
    You're welcome. Sorry again for my poor manners earlier.

    Chet
     
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