How many calories of heat energy are required to completely evaporate 6 g of liquid water?

Question:

Based on the plot below, how many calories of heat energy are required to completely evaporate 6 g of liquid water that is originally 0°C?

a) 12 calories

b) 60 calories

c) 540 calories

d) 660 calories

To vaporize 6 grams of water at 100°C, 3240 calories of heat energy would be required, which is calculated by multiplying 6 grams by the heat of vaporization (540 cal/g). The provided options do not include the correct answer.

Answer:

The quantity of heat energy required to completely evaporate 6 grams of liquid water that is initially at 0°C can be found using the heat of vaporization of water, which is given as 540 calories per gram. So, the total heat energy needed to vaporize 6 grams of water is simply 6 g multiplied by 540 cal/g.

6 g × 540 cal/g = 3240 calories

However, this question may contain an error as none of the given options (12 calories, 60 calories, 540 calories, 660 calories) match the calculated value. If the original question intended to ask about the evaporation of water at 100°C, the answer would be 6 g × 540 cal/g = 3240 calories. But for water at 0°C, there is an additional step which involves heating the water from 0°C to 100°C before it can evaporate, which is not considered in the provided options. Therefore, the correct option is not listed among the choices provided.