Chemistry

A quantity of ice at 0.0°C was added to 25.6 g of water at 21.0°C to give water at 0.0°C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g · °C).

I don't understand how to set this problem up (what the formula is and where to plug in which qualities). Any help would be very much appreciated.

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asked by Ryan
  1. How much heat must be removed from the 25.6 g H2O @ 21.0.0 C to move it to 0.0 C.
    That will be
    q = heat to be removed = mass x specific heat water x delta T.
    q = 25.6 g x 4.18 x 21.0 = about 2000 J (but you need to do it exactly).
    Then 2000 J x 1 g/heat fusion = grams ice.
    You are given the heat of fusion in kJ/mol. You must change that to J/g so it will fit the remainder of the problem. It's approximately 334 J/g but you need to verify that. Post your work if you get stuck.

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    posted by DrBob222
  2. it worked perfectly, thank you so much

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    posted by Ryan

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