Two kilograms of water at 100°C is converted to steam at 1 atm. Find the change in internal energy (in J). (L_v = 2.26 × 10⁶ J/kg.) A) 2.1 × 10⁴ B) 4.5 × 10⁶ C) 3.4 × 10⁵ D) 4.2 × 10⁶ E) 2.1 × 10⁶

Question

Quizplus · Accepted Answer

The change in internal energy can be calculated using the formula:
ΔU = m*L + m*C*ΔT
where m is the mass, L is the specific latent heat of vaporization, C is the specific heat capacity, and ΔT is the change in temperature.
Here, m = 2 kg, L = 2.26 × 10^6 J/kg (specific latent heat of vaporization of water), and C = 4.18 × 10^3 J/kg°C (specific heat capacity of water).
The initial temperature is 100°C, and the final temperature is 100°C (since the water is converted to steam at the same temperature).
Thus, ΔT = 0. Therefore, the change in internal energy is:
ΔU = m*L = 2*2.26 × 10^6 = 4.52 × 10^6 J
Rounding it off to two significant figures gives: 
ΔU ≈ 4.2 × 10^6 J
Therefore, option D is the correct answer.

Two Kilograms of Water at 100°C Is Converted to Steam