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Specific heat problem simple, but im missing something ?

  1. Apr 5, 2009 #1
    specific heat problem...simple, but im missing something ????

    1. The problem statement, all variables and given/known data
    Trying to beat the heat of the last summer, a physics grad student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, he filled it with 200 liters of water at 25 C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0C.) He continued to add ice cubes, until the temperature stabilized to 16C. He then got in the pool.

    The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g C, the specific heat of ice is 0.5 cal/g C, and the latent heat of fusion of water is 80 cal/g.

    How many ice cubes did he add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)



    2. Relevant equations
    m=rho*V
    Q=mc delta T
    QL = m Lf


    3. The attempt at a solution

    volume of water: 200L = .2m^3
    m = rho*V
    m = 1000 * .2 = 200kg of water

    Q of water = mc delta T
    = .2 * 4.186 * 9
    = 7534.8J

    Q of ice = mc delta T
    = .03kg * 2.09 * 16
    = 1.00J

    latent heat of fusion for each cube of ice is..
    QL = m*Lf
    = .03 * (80*4.186)
    = 10.05J

    so Qwater / (Qice + QL) = number of ice cubes.

    7534.8J / (1.00J + 10.05J)

    = 681.88 cubes of ice

    = wrong..

    ??please help

    (i did notice that this answer is intuitively wrong, but cant see the prob??)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 5, 2009 #2
    Re: specific heat problem...simple, but im missing something ????

    That was my first answer as well, but the answer lies in trying to use everything that you are given. Consider Q of ice and the heat of fusion. What is the heat of fusion? What does that heat go into doing? See if that helps you.
     
  4. Apr 6, 2009 #3
    Re: specific heat problem...simple, but im missing something ????

    Thanks for the response.

    Q of ice is the heat that flows into each icecube to raise its temp from 0 to 16deg
    heat of fusion is the heat required to melt an ice cube of mass 30g.
    The heat is used first to melt the ice cube (no temp difference)
    Then it is used to raise the temp to 16deg
    Im using all the info thats given, I still dont get it???
     
  5. Apr 6, 2009 #4
    Re: specific heat problem...simple, but im missing something ????

    someone please help. Ive done all the work, just missing some small thing and dont know what the prob is????
     
  6. Apr 6, 2009 #5

    Redbelly98

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    Re: specific heat problem...simple, but im missing something ????

    The ice cubes are initially at 0C. For the 16C rise, what phase (ice or liquid?) is appropriate? This determines the value of c.
     
  7. Apr 6, 2009 #6
    Re: specific heat problem...simple, but im missing something ????

    for the 16deg rise, the phase that is appropriate is water (im not sure i understad the question?)
    at zero deg it is ice, so i use the latent heat of fusion value that i am given (80cal/g)
    because the ice has to first melt.
    then i multiply this by the ice cube mass (.03kg) and 4.186 (to convert to joules)
    = 10.05J

    then the just melted ice (now water) rises in temp from 0 to 16.
    this is all above.
    i still dont see the error????
     
  8. Apr 6, 2009 #7

    Redbelly98

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    Re: specific heat problem...simple, but im missing something ????

    Okay. You originally used 2.09, the specific heat of ice, instead of 4.186.
     
  9. Apr 6, 2009 #8
    Re: specific heat problem...simple, but im missing something ????

    can you tell me exactly where I am wrong, i dont understand.

    Q of water = mc delta T
    = .2 * 4.186 * 9
    = 7534.8J

    Q of ice = mc delta T
    = .03kg * 2.09 * 16 ( i am using 2.09 because specific heat of ice is 0.5cal/g. Therefore .5*4.186 = 2.09)
    = 1.00J

    latent heat of fusion for each cube of ice is..
    QL = m*Lf
    = .03kg * (80*4.186)
    = 10.05J

    so Qwater / (Qice + QL) = number of ice cubes.

    7534.8J / (1.00J + 10.05J)
    = 681.88 cubes of ice (this is wrong)
     
  10. Apr 6, 2009 #9

    Redbelly98

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    Re: specific heat problem...simple, but im missing something ????

    This is where the mistake is. The ice melts first into liquid water, and then increases by 16 C in the liquid phase.

    Use c = 4.18 J/g/C for liquid water, not 2.09.

    Here you were implying that solid ice changes temperature from 0C to 16C.

     
  11. Apr 6, 2009 #10
    Re: specific heat problem...simple, but im missing something ????

    oooohhh. that was stupid. yes, thanks for the conceptual explanation. that totally makes sense.
    answer = 624.8 cubes of ice.
     
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