How much heat (in BTU) would be required to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm? Assume heat capacity is constant at 1 BTU/lb-°F and the latent heat of vaporization is 1,000 BTU/lb. Also, what is the temperature of the saturated steam at 1 atm?
Is that 25C or 25F? I will assume it is F but if not you can change from F to C and the reverse.
q1 = heat to move water from 25 F to 212F.
q1 = 100 lb x 1 BTU/lb*F x (Tfinal-Tinitial) where Tf-Ti = 212-25
q2 = heat to vaporize water at 212 F to steam at 212 F.
q2 = 100 lb x 1000 BTU/lb.
Total Q = q1 + q2.
The second question answer is in my q2 statement.
To calculate the heat required to heat and vaporize 100 lb of water, we need to consider two processes:
1. Heating the water from 25°C to its boiling point (100°C at 1 atm).
2. Vaporizing the water at its boiling point.
First, let's calculate the heat required to raise the temperature of 100 lb of water from 25°C to 100°C:
Heat = Mass × Specific Heat × ΔTemperature
The mass is given as 100 lb, the specific heat is 1 BTU/lb-°F, and the change in temperature is:
ΔTemperature = 100°C - 25°C = 75°C = 75°F
Heat = 100 lb × 1 BTU/lb-°F × 75°F = 7500 BTU
Next, let's calculate the heat required to vaporize the water. The latent heat of vaporization is given as 1000 BTU/lb, and we have 100 lb of water:
Heat = Mass × Latent Heat of Vaporization
Heat = 100 lb × 1000 BTU/lb = 100,000 BTU
The total heat required to heat and vaporize 100 lb of water is the sum of these two heats:
Total Heat = Heat to Raise Temperature + Heat to Vaporize
Total Heat = 7500 BTU + 100,000 BTU = 107,500 BTU
Therefore, it would require 107,500 BTU to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm.
To find the temperature of saturated steam at 1 atm, you can reference a steam table or use the steam saturation curve. At 1 atm, the saturated steam temperature is approximately 100°C or 212°F.
To calculate the amount of heat required to heat and vaporize water, we need to consider two steps: heating the water from 25°C to its boiling point and then vaporizing it.
First, let's calculate the heat required to heat the water from 25°C to its boiling point:
The heat equation is:
Q = m * c * ΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
Given:
m (mass of water) = 100 lb
c (specific heat capacity of water) = 1 BTU/lb-°F
ΔT (temperature change) = boiling point - initial temperature
The boiling point of water at 1 atm is 100°C or 212°F. So, ΔT = 212°F - 25°C.
Now, we can calculate the heat required to heat the water:
Q_heat = m * c * ΔT
Q_heat = 100 lb * 1 BTU/lb-°F * (212°F - 25°C)
Q_heat = 100 lb * 1 BTU/lb-°F * (212°F - 77°F)
Q_heat = 100 lb * 1 BTU/lb-°F * 135°F
Q_heat = 13,500 BTU
So, it requires 13,500 BTU to heat the water from 25°C to its boiling point.
Next, let's calculate the heat required to vaporize the water:
The heat equation for vaporization is:
Q_vaporization = m * L
where Q_vaporization is the heat energy required for vaporization, m is the mass, and L is the latent heat of vaporization.
Given:
m (mass of water) = 100 lb
L (latent heat of vaporization) = 1,000 BTU/lb
Q_vaporization = m * L
Q_vaporization = 100 lb * 1,000 BTU/lb
Q_vaporization = 100,000 BTU
So, it requires an additional 100,000 BTU to vaporize 100 lb of water.
To find the temperature of the saturated steam at 1 atm, we can refer to a steam table. At 1 atm, the saturated steam temperature is approximately 100°C or 212°F.
Therefore, the temperature of the saturated steam at 1 atm is approximately 100°C or 212°F.