how many calories to change 100 grams of water at 100 degrees C. to 100 grams of steam at 100 degrees C.?

The heat of vaporization of H2O is 540 calories per gram.

So, multiply 540 cal/g by 100 g.

thank you drwls!!

To change 100 grams of water at 100 degrees Celsius to 100 grams of steam at 100 degrees Celsius, you need to calculate the amount of heat required.

The specific heat capacity of water is 4.18 Joules/gram·degree Celsius, and the heat of vaporization for water is 2260 Joules/gram.

First, calculate the heat required to raise the temperature of 100 grams of water from 100 degrees Celsius to its boiling point (100 degrees Celsius):

Q1 = mass × specific heat capacity × (final temperature - initial temperature)
= 100 grams × 4.18 Joules/gram·degree Celsius × (100 degrees Celsius - 100 degrees Celsius)
= 0 Joules

Since the initial and final temperatures are the same, no heat is required to raise the temperature of water to its boiling point.

Next, calculate the heat required to vaporize 100 grams of water at its boiling point:

Q2 = mass × heat of vaporization
= 100 grams × 2260 Joules/gram
= 226,000 Joules

Therefore, to change 100 grams of water at 100 degrees Celsius to 100 grams of steam at 100 degrees Celsius, you need 226,000 Joules of heat.

To convert the heat from Joules to calories, divide by 4.184:

Calories = Joules / 4.184
= 226,000 / 4.184
≈ 54,100 calories

So, it requires approximately 54,100 calories to change 100 grams of water at 100 degrees Celsius to 100 grams of steam at 100 degrees Celsius.

To calculate the number of calories required to change 100 grams of water at 100 degrees Celsius to 100 grams of steam at 100 degrees Celsius, we need to consider the heat required for each phase change separately.

First, we need to heat the water from its initial temperature of 100 degrees Celsius to its boiling point, which is also 100 degrees Celsius. To do this, we can use the specific heat capacity of water, which is approximately 1 calorie per gram per degree Celsius. Therefore, the number of calories required to heat 100 grams of water from 100 to 100 degrees Celsius is:

Calories = mass (grams) * specific heat capacity * temperature change
Calories = 100 g * 1 cal/(g °C) * (100 - 100) °C
Calories = 0

As the temperature does not change when the water is boiling, no calories are required for this phase change. However, the water absorbs heat during the phase change from liquid to gas, known as the latent heat of vaporization.

The latent heat of vaporization for water is approximately 540 calories per gram. Therefore, we can calculate the number of calories required for this phase change as follows:

Calories = mass (grams) * latent heat of vaporization
Calories = 100 g * 540 cal/g
Calories = 54,000

So, to change 100 grams of water at 100 degrees Celsius to 100 grams of steam at 100 degrees Celsius, it would require a total of 54,000 calories.