Calculate the increase in the internal energy when 1000 g of boiling water at 212° F is converted into steam at the same temperature. Atmospheric pressure is 1.013 X 10 Pa and the latent heat of vaporization is 2.260 MJ kg
-1

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.

i think the answer would be b