Calculate the amount of heat required to change 5 pounds of 212°F water to steam
I'm not an engineer so I don't know the formula with correct symbols but
heat = mass H2O x delta Hvaporization @ 212 F
heat = 5 lb x 970 BTU/lb = ?BTU
Delta H vaporizatiion is also called the latent heat of vaporization.
To calculate the amount of heat required to change water to steam, you need to consider two steps: heating the water from its initial temperature to the boiling point, and then converting the water at boiling point to steam.
Step 1: Heating water from 212°F to boiling point:
The specific heat capacity of water is 1 calorie/g°C. Therefore, it takes 1 calorie of heat to raise the temperature of 1 gram of water by 1 degree Celsius (1 calorie/g°C).
First, convert the weight of water from pounds to grams:
5 pounds = 5 * 453.592 grams (1 pound = 453.592 grams)
= 2267.96 grams
To calculate the heat required, use the formula:
Q = m * c * ΔT
Where:
Q = heat required
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Q1 = 2267.96 g * 1 calorie/g°C * (100°C - 212°F)
Note: The conversion from Fahrenheit to Celsius is (°F - 32) * 5/9.
Q1 = 2267.96 g * 1 calorie/g°C * (100°C - (212°F - 32) * 5/9)
Q1 = 2267.96 g * 1 calorie/g°C * (100°C - (180 * 5/9))
Q1 = 2267.96 g * 1 calorie/g°C * (100°C - 100°C)
Q1 = 2267.96 g * 1 calorie/g°C * 0°C
Q1 = 0 calories
Therefore, no heat is required to heat water from 212°F to its boiling point.
Step 2: Converting water to steam at boiling point:
The heat required to convert water at its boiling point to steam is called the latent heat of vaporization. For water, the latent heat of vaporization is approximately 540 calories/gram.
Q2 = m * L
Where:
Q2 = heat required
m = mass of water
L = latent heat of vaporization
Q2 = 2267.96 g * 540 calories/gram
Q2 = 1,223,558.4 calories
Therefore, the amount of heat required to change 5 pounds of 212°F water to steam is approximately 1,223,558.4 calories.