Question

A quantity of ice at 0.0 °C was added to 33.6 g of water at 41.0 °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). ______ grams

Answer 1

Heat lost by water = heat gained by ice

                mcdt = m'L

Where

m = mass of water = 33.6 g

c = specific heat = 4.18 J/(g•°C)

dt = change in temperature of water = final - initial = 41.0 - 0 = 41.0 ^oC

m' = mass of ice = ?

L = heat of fusion of water = 6.01 kJ/mol

                                      = 6.01x10³ J/mol

                                      = 6.01x10³ (J/mol) x (1/18)( mol /g)

                                      = 334 J/g

Plug the values we get

mcdt = m'L

33.6x4.18x41 = m'x334

m' = 17.2 g

Therefore the quantity of ice added is 17.2 grams