Answer :
The total energy required to convert 5 pounds of water at 212°F to steam at 216°F is approximately 5137.57 kJ.
To find out how much energy is needed to convert 5 pounds of water at 212°F to steam at 216°F, we need to consider two main steps: heating the water to steam (phase change) and then heating the steam further.
Phase Change (Water to Steam at 212°F/100°C):
The latent heat of vaporization of water is around 2256 kJ/kg.
5 pounds of water is equivalent to approximately 2.27 kg (using 1 pound = 0.453592 kg).
The energy required for the phase change is calculated as:
Q₁ = mL
Where:
Q₁ is the heat energy required for the phase change
m is the mass (2.27 kg)
L is the latent heat of vaporization (2256 kJ/kg)
Therefore,
Q₁ = 2.27 kg * 2256 kJ/kg = 5127.12 kJ
Heating Steam from 212°F (100°C) to 216°F (102.22°C):
The specific heat capacity of steam is approximately 2.08 kJ/kg°C.
The temperature change (ΔT) is from 100°C to 102.22°C (which is 2.22°C).
The energy required to heat the steam is calculated as:
Q₂ = mcΔT
Where:
Q₂ is the heat energy required to raise the temperature
m is the mass (2.27 kg)
c is the specific heat capacity (2.08 kJ/kg°C)
ΔT is the temperature change (2.22°C)
Therefore,
Q₂ = 2.27 kg * 2.08 kJ/kg°C * 2.22°C = 10.45 kJ
Finally, we add both energies to find the total energy required:
Q total = Q1 + Q2
Therefore,
Q total = 5127.12 kJ + 10.45 kJ = 5137.57 kJ
