how many kilocalories of heat are released when 85g steam at 100 degrees C converted to ice at 0.

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  1. Three steps are involved: steam changing to water at 100 C; water at 100 C cooling to 0 C; water at 0 C freezing to ice at 0 C.
    Steam changing to water:
    Heat of vaporization of water is 539.4 cal/g
    Heat = Heat of Vap. x mass = 539.4 cal/g x 75.0 g = 40,500 cal
    Cooling water from 100 C to 0 C:
    Heat = mass x specific heat x T change
    Heat = 75.0 g x 1.00 cal/gC x 100 C = 7500 cal
    Freezing of water at 0 C:
    The heat of fusion of water is 79.7 cal/g
    Heat = heat of fusion x mass = 79.7 cal/g x 75.0 g = 5980 cal
    Total heat = 40,500 + 7500 + 5980 = 53980 cal or 5.40 x 10^4 cal (3 sig figs) = 54.0 kcal

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    posted by Elina
  2. The work above is correct except the process is a bit confusing and hard to follow. I teach this process to my Honors Chemistry students in a similar way, just try to start with your grams and use factor labeling for the easiest way to solve.

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