Final answer:
To convert 3 grams of ice at 0 degrees Celsius to water vapor at 100 degrees Celsius, 240 calories are required for melting, 300 calories for heating, and 1620 calories for vaporization, totaling 2160 calories.
Step-by-step explanation:
To answer the question about how many calories are required to convert 3 grams of ice at 0 degrees Celsius to water vapor at 100 degrees Celsius, we need to consider the heat of fusion, the specific heat capacity of water, and the heat of vaporization.
First, to melt the ice into water, we use the heat of fusion. Since 1 gram of ice requires 80 calories (or 334 J/g as provided in Question 15A), 3 grams of ice would require:
- 3 g × 80 cal/g = 240 calories
Next, to heat the water from 0 to 100 degrees Celsius, we use the specific heat capacity of water, which is 1 cal/g°C. Thus:
- 3 g × 100°C × 1 cal/g°C = 300 calories
Finally, to vaporize the water into steam, we use the heat of vaporization, which is 540 calories per gram. Therefore, for 3 grams:
- 3 g × 540 cal/g = 1620 calories
Adding these together:
- 240 cal (fusion) + 300 cal (heating) + 1620 cal (vaporization) = 2160 calories
The heat is positive because the process is endothermic; energy is absorbed to break the molecular bonds during the phase changes from solid to liquid and from liquid to gas.