What is the thermal energy needed to completely vaporize 18.02g or water at 100 degrees celcius?

Question

a. 6.01 kJ

b. 40.7 kJ

c. 6.01 x 10^3 kJ

d. 4.07 x 10^4 kJ

Answer 1

molecular weight of water = 16.016 g/mol

heat of vaporization of water = 40.650 kJ/mol

heat = (40.650)(18.02) / (16.016)

= 45.736 kJ

closer to (b)

Answer 2

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