If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? (Assume the density of water is 1.00 g/mL.)

0.425 L

Explanation:

Given the information from the question. We know that the water heat evaporation is given by 2261 KJ/kg. Therefore, for each Kg of vapor, 2261 is required.

To calculate the amount of water evaporated by 960 KJ is as follows

= (1/2260) * 960 = 0.425 Kg

In other words, 0.425 of water is evaporated by 960 KJ.

To calculate the volume of water, we have mass and density. Therefore, Volume = Mass/Density = 425g/1g/ml = 0.425 L.

Therefore, water evaporated is 0.425 L