A quantity of ice at 0.0°C was added to 25.0 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).

Question

A quantity of ice at 0.0°C was added to 25.0 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).
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DrBob222 · Answer

(g ice x heat fusion) +[g melted ice x specific heat H2O x (Tfinal-Tinitial)] + [g H2O x specific heat water x (Tfinal-Tinitial)] = 0
You have only one unknown namely g ice. Solve for that. I estimated the answer to be about 7g ice at zero. Note that I have three terms above; i.e., ice + melted ice + water at 25. The middle term (melted ice) is multiplied by (Tfinal-Tinitial) but since Tfinal is zero and Tinitial is zero then Tf-Ti = 0 and anything times zero is zero so the middle term is zero.