How many calories are needed to change 1 gram of 100 degrees Celsius boiling water to 100 degrees Celsius steam?
To calculate the amount of calories needed to change 1 gram of boiling water to steam at 100 degrees Celsius, we need to consider the energy required for each phase change and the specific heat capacities involved.
1. Energy to heat water from 100 degrees Celsius to its boiling point:
The specific heat capacity of water is approximately 1 calorie/gram°C. Therefore, the energy required to heat the water from 100 degrees Celsius to its boiling point, which is 100 degrees Celsius, can be calculated as:
Energy = mass * specific heat capacity * temperature difference
Energy = 1 gram * 1 calorie/gram°C * (100 - 100) degrees Celsius
Energy = 0 calories
There is no energy required to reach the boiling point as the temperature remains constant during this phase change.
2. Energy for phase change from liquid (water) to gas (steam):
The heat of vaporization, which is the energy required to convert a substance from liquid to gas at its boiling point, is approximately 540 calories/gram for water. Therefore, the energy required for this phase change is:
Energy = mass * heat of vaporization
Energy = 1 gram * 540 calories/gram
Energy = 540 calories
Therefore, to change 1 gram of 100 degrees Celsius boiling water to 100 degrees Celsius steam, approximately 540 calories of energy are required.
To determine the number of calories needed to change 1 gram of boiling water to steam at 100 degrees Celsius, we need to consider the specific heat capacity and latent heat of vaporization.
1. First, we need to calculate the calories required to raise the temperature of water from 100 degrees Celsius to its boiling point.
The specific heat capacity of water is approximately 1 calorie/gram °C. Therefore, the calories required to raise the temperature by 1°C is:
1 gram × 1 calorie/gram °C = 1 calorie
Since we need to raise the temperature by 100°C, the total calories required for this step is:
1 calorie/°C × 100°C = 100 calories
2. Once the temperature of water reaches the boiling point, additional energy is required for a phase change from liquid water to steam, known as latent heat of vaporization.
The latent heat of vaporization for water is approximately 540 calories/gram.
Therefore, the total calories required for the phase change is:
1 gram × 540 calories/gram = 540 calories
Adding the calories required for both steps:
100 calories (for temperature increase) + 540 calories (for phase change) = 640 calories
Hence, it would take approximately 640 calories to change 1 gram of boiling water at 100 degrees Celsius to 1 gram of steam at 100 degrees Celsius.