Question

How much heat is absorbed to change 13.8 grams of H2O from a solid to a liquid at zero degrees Celsius? ΔHfusion = 6.03 kJ/mol ΔHvaporization = 40.65 kJ/mol
31.1 kJ

561 kJ

4.62 kJ

83.2 kJ

Answer 1

1 mole of H2O weighs 18 g

therefore 13.8 g of liquid H2O = 13.8/18 moles

ΔHvaporization = 40.65 kJ/mol

heat required to change 13.8 grams of H2O from a liquid to a gas at 100 degrees Celsius = 40.65 x 13.8/18 = 31.165 kJ

Answer 2

Answer:

4.62kJ

Explanation:

The heat is required for

Q1 = Heat required for conversion of solid to liquid

As there is no change in temperature so no heat will be absorbed in increasing the temperature.

Q1 = moles of water X heat of fusion of water

$moles=\frac{mass}{molarmass}=\frac{13.8}{18}=0.77mol$

Q1= 0.7667 X6.03kJ/mol = 4.62kJ