Final answer:
To calculate the calories required to heat and vaporize 100 g of liquid water from 25 °C to 100 °C, we add the heat needed to raise the temperature to 100 °C (7,500 cal) to the heat of vaporization (54,000 cal), resulting in a total of 61,500 calories.
Step-by-step explanation:
To determine the calories required to heat 100 g of liquid water at 25 °C to water vapor at 100 °C, we need to perform two calculations: one for heating the water from 25 °C to 100 °C, and one for vaporizing the water at 100 °C.
Firstly, to heat the water to its boiling point, we use the specific heat capacity of water which is 1 calorie per gram per degree Celsius (1 cal/g°C). Therefore, the heat required to raise the temperature of 100 g of water by 75 °C (from 25 °C to 100 °C) is:
Heat = (mass) x (specific heat capacity) x (temperature change)
Heat = (100 g) x (1 cal/g°C) x (100 °C - 25 °C)
Heat = (100 g) x (1 cal/g°C) x (75 °C)
Heat = 7,500 calories
Secondly, the heat of vaporization of water is 540 calories per gram. To vaporize 100 g of water at 100 °C, the heat required is:
Heat = (mass) x (heat of vaporization)
Heat = (100 g) x (540 cal/g)
Heat = 54,000 calories
Adding both amounts together gives us the total heat required to heat and vaporize 100 g of water from 25 °C to water vapor at 100 °C:
Total Heat = Heat to raise temperature + Heat to vaporize
Total Heat = 7,500 cal + 54,000 cal
Total Heat = 61,500 calories
The correct answer is: (c) 61,500 cal.