Homework Answers
Amount of heat required to raise the temperature from 25°C to 100°C (boiling point of water):
q1 = mc delta t
Given:
m = 14g
c = 4.18J/gC
delta t = (100-25)C = 75°C
q1 = 14×4.18×75 = 4389J = 4.389kJ
Latent heat needed to change liquid water to vapour
q2 = mL
L = heat of vapourization = 40.7kJ/mol
Moles of water = 14g/18g/mol = 0.77mol
q2 = 0.77mol × 40.7kJ/mol = 31.65kJ
Total heat = q1 + q2 = 4.389 + 31.65 = 35.85kJ
Heat required = 35.85kJ
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