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Homework Help: Cooling Off (Heat, Temperature and Phase Changes)

  1. Jun 23, 2008 #1
    Okay, so I've approached this problem from a number of different ways and have gotten some crazy answers. The poor student would be spending the day in a bath of ice if my answers were right :P

    Here's the problem :

    Trying to beat the heat of the last summer, a physics student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, she filled it with 200 liters of water at 25oC. Realizing that the water would probably not be cool enough, she threw ice cubes from her refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0oC.) She continued to add ice cubes, until the temperature stabilized to 16oC. She 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 she add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

    Number of Ice Cubes = cubes

    HELP: Heat lost by water = heat gained by ice cubes. No heat is lost to the surroundings.

    HELP: Since the water (subsystem 1) is at a higher temperature, heat will be lost to the ice cubes (subsystem 2). Calculate the heat H that the water gave up from 25 C to 16 C; calculate the heat h that each ice cube gained from 0C to 16 C including melting. Then the number of ice cubes equals H/h.

    So I tried to follow what was given in the help section, determining the Heat Lost by the water and that gained by an ice cube. I then set them up with Qw / Qice as suggested but keeping getting the wrong answer. Any help would be appriciated!

    My Work :

    http://img206.imageshack.us/img206/620/workjs0.jpg [Broken]

    BTW - I used the standard values for specific heat and latent heat, but I don't believe that should make much of a difference.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 23, 2008 #2


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    What happens to the ice when it has melted? Does it stay at 0oC?
    Last edited by a moderator: May 3, 2017
  4. Jun 23, 2008 #3
    Oh, it continues to heat up?

    So I need to add 0.03kg(4186)(16-0) to the demoninator? But it still comes out to the wrong answer. I get 625 ice cubes.

    I did 7534800 / (10050 + 2009.28) <--- Original demoniator plus the energy from continued heating.
  5. Jun 23, 2008 #4
    Oh I figure out my mistake, I wasn't doing 0.03kg * 2090 because I thought since the change in T was zero it wasn't significant. I guess the student really is going to spend the time adding 621 ice cubes o_0

    Thanks for all the help! :)
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