Ten pounds of liquid water is at a temperature of 180 degrees F at standard atmospheric conditions.
Determine the energy required to convert this to water vapor (steam) at a temperature of 212 degrees F.
Do you want this in BTU?
q1 = heat to move from 180 to 212.
q1 = mass x specific heat x (212-180)
q2 = heat to vaporize water at 212 to steam at 212 F.
q2 = mass x heat vaporization.
Total heat = q1 + q2.
I don't know how to find the specific heat in q1.
I have q1 = 320 (specific heat)
#2) How do I find heat vaporitzation for q2?
You don't have a table in your text/notes?
specific heat H2O = 1.00 BTU/lb*F
heat vaporization = 970.4 BTU/lb
To determine the energy required to convert liquid water to steam, we need to calculate the heat energy needed for the phase change from liquid to vapor.
The energy required for this phase change can be calculated using the formula:
Q = m * L
Where:
Q is the energy required
m is the mass of the substance
L is the latent heat of vaporization
First, let's calculate the mass of water. Given that we have 10 pounds of liquid water, we need to convert it to mass. Since 1 pound is equivalent to approximately 0.454 kg, we can convert the mass as follows:
m = 10 pounds * 0.454 kg/pound
m = 4.54 kg
Next, we need to determine the latent heat of vaporization for water. The latent heat of vaporization for water is approximately 2260 kJ/kg, which means that it takes 2260 kJ of energy to convert 1 kg of liquid water into steam.
Now we can calculate the energy required:
Q = m * L
Q = 4.54 kg * 2260 kJ/kg
Q ≈ 10,264.4 kJ
Therefore, the energy required to convert 10 pounds of liquid water to steam at the given conditions is approximately 10,264.4 kilojoules.