To determine the change in internal energy of the water when it is converted to steam at 100ºC and a constant pressure of 1.0087 x 10^5 Pa, we can use the formula:
ΔU = m * Lv
where ΔU is the change in internal energy, m is the mass of the water, and Lv is the latent heat of vaporization.
Given:
m = 0.2 kg
Lv = 2.26 x 10^6 J/kg
Using the formula, we can calculate:
ΔU = 0.2 kg * 2.26 x 10^6 J/kg
ΔU = 452,000 J
Therefore, the change in internal energy of the water when it is converted to steam at 100ºC and a constant pressure of 1.0087 x 10^5 Pa is 452,000 J.
A cylinder with a moveable piston contains 0.2 kg of water at 100 ºC.
The density of the water ρw
=1000 kg/m3 and that of steam is ρs
=0.6
kg/m3
. The Latent heat of vaporization of water is Lv
=2.26x106 J/kg.
Determine the change in internal energy of the water when it is
converted to steam at 100 ºC at a constant pressure of 1.0087 x 105
Pa
1 answer