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:
C
Explanation:
Melting point, temperature at which the solid and liquid forms of a pure substance can exist in equilibrium. As heat is applied to a solid, its temperature will increase until the melting point is reached. More heat then will convert the solid into a liquid with no temperature change.
It is equal because they both weight the same