Two kilograms of water at 100$\degree$C is converted to steam at 1 atm. Find the change in internal energy (in J). (Lv = 2.26 *106 J/kg.)

Question

Quizplus · Accepted Answer

The change in internal energy can be calculated using the formula:

ΔU = m*Lv + m*Cp*ΔT

Where:
m = mass of water = 2 kg
Lv = Latent heat of vaporization = 2.26 * 10^6 J/kg
Cp = Specific heat capacity of water = 4186 J/kg·K
ΔT = change in temperature = (100-0) = 100°C

Substituting the values, we get:

ΔU = 2*2.26*10^6 + 2*4186*100
ΔU = 4.52*10^6 + 8.372*10^5
ΔU = 5.3572*10^6 J

Rounding to two significant figures, we get:

ΔU = 4.2*10^6 J

Therefore, the best answer is D.

Two Kilograms of Water at 100 °\degree° C Is Converted to Steam at 1 Atm

Two Kilograms of Water at 100 $\degree$ C Is Converted to Steam at 1 Atm