Molten iron at 1538 degree C (its melting point) is solidified by an ice-water m

Question

Molten iron at 1538 degree C (its melting point) is solidified by an ice-water mixture at 0 degree C. Assuming that there is still ice present when the process is over and the iron is left in the ice-water, how much ice will be melted if 10.0 g of iron is solidified? The following may be useful: enthalpy of fusion of iron 13.8 kJ/mol; enthalpy of fusion of water = 6.07 kJ/mol; specific heat capacity of iron = 0.449 J/g middot K; specific heat capacity of water = 4.184 J/g middot K. 20.5 g 14.7 g 27.8 g 7.35 g 9.38 g

Explanation / Answer

Qwater = -Qiron

Qiron = Qlatent heat + Qsolid

Qlatent heat = n*LH = mass/MW*LH = (10/(55.8)*13.8 = -2.4731 kJ

Qsolid = n*C*(Tf-Ti) = (10/55.8)*0.449*(0-1538) = -123.756 J

Qiron = -2.4731*1000 - 123.756 = -2596.856 J

So..

for water

Qwater = 2596.8 J

Qwater = n*LW

n = 2596.8/(6.07*1000) = 0.4278 mol

mass = mol*MW = 0.4278*18 = 7.7 g

nearest is 4

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