Two kilograms of water at 100 $\degree$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

First, we need to determine the amount of heat required to convert 2 kg of water at 100°F to steam at the same temperature and pressure (1 atm). This can be calculated using the formula:

Q = m*L

where Q is the heat required, m is the mass, and L is the latent heat of vaporization at 100°F, which is given as 2.26 x 10^6 J/kg.

So,

Q = 2 kg * 2.26 x 10^6 J/kg = 4.52 x 10^6 J

Next, we need to calculate the change in internal energy. Since the process is taking place at constant pressure, we can use the formula:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added, and W is the work done. In this case, the work done is zero, since the process is taking place at constant pressure. Therefore,

ΔU = Q = 4.52 x 10^6 J

Rounding to one significant figure gives the answer of D: 4.2 x 10^6 J.

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