Question

How much heat required to convert 36 grams (2 moles) of ice at 0 degrees C to liquid water at 50 degrees C? delta-H of fusion for H2O is 6.02 kJ/mol and the heat capacity of liquid water is 4.18J/g degrees C.

A) 12 kJ
B) 7524 J
C) 8728 J
D) 19564 J

Answer 1

Mass of ice = 36 g.

Molar mass of ice = 18 g/mol.

Mol(s) ice corresponding to 36 g = (36 g)/18 g/mol) = 2 mol.

The heating process consists of the following sub-processes.

i) Melting of ice to water at 0ºC = (mols ice)*(ΔH_fusion, ice)

= (2 mols)*(6.02 kJ/mol)

= 12.04 kJ

= (12.04 kJ)*(1000 J)/(1 kJ)

= 12040 J.

ii) Heating water from 0ºC to 50ºC

= (mass of water)*(heat capacity of water)*(change in temperature of water)

= (36 g)*(4.18 J/g.ºC)*(50ºC - 0ºC)

= (36 g)*(4.18 J/g.ºC)*(50ºC)

= 7524 J.

Total heat required = (12040 + 7524) J

= 19564 J.

The correct answer is (D) 19564 J (ans).