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Heat capacity / letent heat question

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    An ice cube of mass 40g is transferred directly from the freezer where it is at -8 CELSIUS to a cup containing 120g of water at 95 degrees CELSIUS. Calculate the final temperature of water. Ignore any heat transfer to the cup or its surroundings

    2. Relevant equations
    tf = ti + delta t

    Is there a possibility elimination/substitution is involved? :S



    3. The attempt at a solution

    0.04 (2.1 x 10^3)(delta t) + (0.12)(4.18 x 10^3)(delta t)

    I solved for it and its wrong. The teacher gave us the answer, but not how to solve it so we can check it later. The answer is 50 degrees Celsius. Can someone show me how they solved it? :S
     
    Last edited: Nov 10, 2011
  2. jcsd
  3. Nov 10, 2011 #2

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    Welcome to PF, blarginsnarf! :smile:

    You are skipping a few steps here.
    In particular you have not included the latent heat.

    Let's start with your "delta t".
    There is not just one "delta t" here, but there are 3 delta-t's.

    First you have the delta t for the ice from its initial temperature to the melting point.
    Second you have the delta t for the ice from the melting point to the as yet unknown final temperature.
    And third you have the delta t for the hot water from its initial temperature to the unknown final temperature.

    Can you write these 3 delta-t's in formula form?
     
  4. Nov 10, 2011 #3
    Alright, this is what I got:
    1. delta t= 0 - (- 8)
    delta t = 8
    2. delta t = tf - 0
    3. delta t = tf - 95


    How's that? :S
     
  5. Nov 10, 2011 #4

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    Good! :)

    What would be the heat absorbed by the ice (in formula form) to bring it first to the melting point, then melt it, and then bring it up to the final temperature?
     
  6. Nov 10, 2011 #5
    Well according to my textbook, the heat capacity of ice is: 2.1 x 10 ^ 3

    So... I did this: What would be the heat absorbed by the ice (in formula form) to bring it first to the melting point:

    q = mct
    q = 0.04 x 2.1 x 10 ^ 3 x 8
    q = 672 J



    then melt it:

    Latent heat formula:
    Latent heat of fusion for water in my textbook = 3.4 x 10 ^ 5

    Qf = mlf
    Qf = 0.04(3.4 x 10 ^ 5)
    Qf = 13600 J




    and then bring it up to the final temperature?:

    I'm not sure at all :S

    What I'm assuming: mass of ice + mass of water * latent heat of fusion for water? :S
     
  7. Nov 10, 2011 #6

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    Yep! :smile:


    Good!


    That would be, with a heat capacity of water of 4.2 x 10 ^ 3:
    q = mcΔt
    q = 0.04 x 4.2 x 10 ^ 3 x (tf - 0)


    Huh? I don't understand what you're saying here.
    You're adding a mass to a heat? :confused:




    Next can you tell me what the heat given off by the hot water is (in formula form)?

    The sum of the heats that the ice absorbs has to be equal to the heat the hot water gives off.
    Can you set up the equation for that?
     
  8. Nov 10, 2011 #7

    Alright, so 672 + 13600 + 168tf
    = 14272 + 168tf? :s I dont know tf


    Next can you tell me what the heat given off by the hot water is (in formula form)?

    I THINK it's like this:

    Q = MCT
    Q = 0.12 x 4.2 x 10 ^ 3 x(tf - 95)
    Q = -46368 + 504tf
    How's that? :D
     
  9. Nov 10, 2011 #8

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    What's not to know?
    That is right!


    Great!

    Now the heat absorbed by the ice must be equal to the heat given off by the hot water.
    Can you set them equal to each other (mind the minus sign) and solve for tf?
     
    Last edited: Nov 10, 2011
  10. Nov 10, 2011 #9
    Oh my gosh, I just solved it on paper. Thank you so much!!! You are a legend!
     
  11. Nov 10, 2011 #10

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