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Calorimetry and Enthalpy Ice Lab

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    This was the procedure for the lab:

    Find the mass of the Styrofoam cup. Half fill it with warm water at a temperature of about 50 degrees C. Determine the mass of the water and the cup. Dry off some small pieces of ice with the paper towel. Record the temperature of the water. Quickly place the ice into the water and stir it until all the ice has melted. Record the lowest temperature obtained by the water. This may be a minute or so after all the ice has melted. Find the final mass so that the mass of the ice added can be found.

    This was the data I collected:

    Initial mass of cup in grams: 2.19g
    Mass of cup with warm water in grams: 140.61 grams
    Final mass of cup with contents in grams: 164.02 grams
    Initial temperature of water: 69 C
    Final temperature of water: 49 C
    Specific heat capacity of water: 4.18 J/g/C

    These are the questions I need to answer:
    1. Calculate the following using q=mct
    a) the heat lost by the warm water
    b) the heat gained by the water formed from the melted ice

    2. With the above two values in mind, how much energy was therefore used to melt the ice?

    3. Knowing the mass of the ice used, calculate the molar heat of fusion of ice in J/g and in J/mol.

    I have no idea how to do numbers 2 and 3.
    Anything helps!!!!

    2. Relevant equations

    I'm just going by what I think I'm supposed to use.

    Q=mct -- to find heats

    Delta H = n * delta H(r) -- I think this is related to molar enthalpy some how??

    Q absorbed + Q released = 0

    3. The attempt at a solution

    1. a) To find the heat lost by the warm water.

    mass = 140.61 - 2.19
    = 138.42 g

    delta t = 49-69
    = -20 degrees celsius

    c = 4.18 J/g degrees Celsius for water

    Q = mct
    = (138.42)(4.18)(-20)
    = -11 571.912 J -- the heat lost by the warm water

    b) Mass of ice water: 164.02 - 140.61 = 23.41 g

    Delta t: 49 - 0 = 0 degrees C

    c = 4.18 J/g/C
    Q = mct
    = (23.41)(4.18)(49)
    = 4794.84 J -- the heat gained by the water formed by the melted ice

    2. Q absorbed + Q gained = 0

    -11 571.912 + 4794.84 = 0
    -6777.08 = 0

    Therefore, the energy needed to melt the ice was 6777.08 J.

    I'm not quite sure if any of these are right...and question 3 doesn't make sense to me.
     
    Last edited: Jan 20, 2014
  2. jcsd
  3. Jan 20, 2014 #2

    SteamKing

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    Staff Emeritus
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    Homework Helper

    It takes a certain amount of heat to turn 1 g of ice at 0 degrees C to 1 g of liquid water at 0 degrees C. This amount of heat is known as the heat of fusion. It appears that you are to calculate this value in joules per gram and also in joules per mole. You do understand that the phase change from solid to liquid occurs at constant temperature?
     
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