1m ^ 3 of water is converted into 1671m ^ 3 of steam at 1 atmosperic pressure and 100 deg * C temperature. The latent heat of vapourisation of water is 2.3 * 10 ^ 6 * J / k * g If 2.0 kg of water is converted into steam at 1 atm pressure and 100 deg * C temperature, how much will be the increase in internal energy? [Density of water = 10 ^ 3 * kq / m * 3 1atm = 1.01 * 10 ^ 5 * N * m ^ - 2 * 1

Answer 1

First, let's calculate the amount of heat required to convert 1m^3 of water into steam using the latent heat of vaporization:

Q = m * L
Q = 1000 kg * 2.3 * 10^6 J/kg = 2.3 * 10^9 J

Now, let's calculate the increase in internal energy for 2.0 kg of water:

ΔU = Q - W
ΔU = 2.3 * 10^9 J - PΔV

Since the volume change is negligible for water, we can ignore the work done:

ΔU = 2.3 * 10^9 J

Therefore, the increase in internal energy for 2.0 kg of water converted into steam at 1 atm pressure and 100 deg C temperature will be 2.3 * 10^9 J.