Question

How much energy is needed to vaporize 25.0 g of H2O at 100°C, given that ΔHvap of water = 40.7 kJ/mol?

A) 20.35 kJ
B) 81.4 kJ
C) 40.7 kJ
D) 10.175 kJ

Answer 1

The amount of energy needed to vaporize 25.0 g of H2O at 100°C is 13.5 kJ.

Step-by-step explanation:

The amount of energy needed to vaporize a given mass of water can be calculated using the equation Qv = mLv, where Qv is the amount of energy, m is the mass of water, and Lv is the heat of vaporization of water. We are given that the heat of vaporization of water is 540 cal/g, so we can convert this to kJ/g by dividing by 1000. To find the amount of energy needed to vaporize 25.0 g of water at 100°C, we multiply the mass by the heat of vaporization:

Qv = (25.0 g)(540 cal/g) = 13,500 cal

To convert this to kJ, we divide by 1000:

Qv = 13,500 cal / 1000 = 13.5 kJ