Question

Calculate the amount of heat needed to melt 2.00 kg of iron at its melting point (1809 K), given that: ?Hfus = 13.80 kJ/mol.

Answer 1

Mass of iron = 2.00 kg = 2000 g

Moles of iron = mass/molar mass

= 2000/55.847 = 35.812 mol

Heat needed = moles x ΔHfus

= 35.812 x 13.80

= 494.2 kJ

Answer 2

Amount of heatrequired , Q = mL
wherem = mass of iron = 2 Kg = 2000 gL = latent heat of fusion of iron = 13.80K/mol= 13800J/molSince 1KJ = 1000 JSubstitute the above values we get Q =mL= 2000 g x 13800 J/mol= 27600000 J= 27600 KJ

Answer 3

Amount of heatrequired , Q = mL
wherem = mass of iron = 2 Kg = 2000 gL = latent heat of fusion of iron = 13.80K/mol= 13800J/molSince 1KJ = 1000 JSubstitute the above values we get Q =mL= 2000 g x 13800 J/mol= 27600000 J= 27600 KJ

Answer 4

Energy needed to change phase is q = moles of substance * heat of fusion

2000 g iron/55.85 g/mol = 35.8 mol iron

Heat of fusion for iron = 13.8 kJ/mol

35.8 mol * 13.8 kJ/mol = 494.1 kJ