Answer:
The increase in the internal energy is 1.840 x 10⁶ J.
Explanation:
Given;
mass of the water, m = 1000 g = 1 kg
temperature of the boiling water, t = 212 ° F = 100 ° C
latent heat of vaporization, L = 2.260 MJ/kg = 2.260 x 10⁶ J/kg
The internal energy of the boiling water is calculated as;
Q₁ = mcΔθ
where;
c is specific heat capacity of water, = 4200 J/kg.⁰C
Δθ is change in temperature = 100 ° C
Q₁ = 1 x 4200 x 100
Q₁ = 420,000 J
The internal energy of the vaporized steam is calculated as;
Q₂ = mL
Q₂ = 1 x 2.260 x 10⁶
Q₂ = 2,260,000 J
The increase in the internal energy is calculated as;
ΔQ = Q₂ - Q₁
ΔQ = 2,260,000 J - 420,000 J
ΔQ = 1,840,000 J
ΔQ = 1.840 x 10⁶ J
Therefore, the increase in the internal energy is 1.840 x 10⁶ J.
