how much heat is absorbed when 63.7 grams of water at 100 degrees Celsius is converted to steam at 100 degrees Celsius?
-I need help figuring out how to do this.
q = mass H2O x heat vaporization
To determine the heat absorbed during the conversion of water to steam, you need to consider two things: the heat required to raise the temperature of the water from 100 degrees Celsius to its boiling point and the heat of vaporization.
Here's how you can calculate it step-by-step:
1. Calculate the heat absorbed to raise the temperature of water from 100 degrees Celsius to its boiling point:
The specific heat capacity of water is typically considered to be 4.18 J/g°C.
Q1 = m * c * ΔT
= 63.7 g * 4.18 J/g°C * (100°C - 100°C)
= 0 J
As there is no temperature change, no heat is absorbed during this step.
2. Calculate the heat absorbed for the phase change from liquid water to steam:
The heat of vaporization for water is 2260 J/g.
Q2 = m * ΔHvap
= 63.7 g * 2260 J/g
= 143,942 J
Therefore, 143,942 J of heat is absorbed during the conversion of water to steam.
Note: It is important to utilize the correct unit conversions and significant figures throughout the calculation process.
To determine how much heat is absorbed when water is converted from 100 degrees Celsius to steam at 100 degrees Celsius, we need to use the formula:
Q = m * ΔH
Where:
Q is the heat energy absorbed/released
m is the mass of water
ΔH is the heat of vaporization (or latent heat), which is the amount of heat required to convert 1 gram of water to steam at its boiling point.
To solve this problem, we need to know the specific heat of water and the heat of vaporization.
1. Determine the heat of vaporization of water:
The heat of vaporization of water is approximately 40.7 kJ/mol. This value can be found in reference books or online sources.
2. Convert the mass of water from grams to moles:
To do this, we need to know the molar mass of water (H2O), which is approximately 18 g/mol.
Moles of water = mass / molar mass = 63.7 g / 18 g/mol = 3.54 moles
3. Calculate the heat energy absorbed:
Plug the values into the formula:
Q = m * ΔH
Q = 3.54 moles * 40.7 kJ/mol = 144.14 kJ
So, approximately 144.14 kJ of heat energy is absorbed when 63.7 grams of water at 100 degrees Celsius is converted to steam at 100 degrees Celsius.