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

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Rewritten by : Barada