First, we need to find the amount of heat required to convert 1m^3 of water into steam.
Given:
Mass of water = density * volume = 10^3 kg/m^3 * 1 m^3 = 10^3 kg
Latent heat of vaporization = 2.3 * 10^6 J/kg
Therefore, the total heat required = mass * latent heat = 10^3 kg * 2.3 * 10^6 J/kg = 2.3 * 10^9 J
Now, for 2.0 kg of water, the amount of heat required would be:
2.0 kg * 2.3 * 10^6 J/kg = 4.6 * 10^6 J
This heat energy is used to convert water into steam and increase the internal energy of the system.
The increase in internal energy is the same as the heat energy required to convert the water into steam:
4.6 * 10^6 J
Therefore, the increase in internal energy when 2.0 kg of water is converted into steam at 1 atm pressure and 100 deg C temperature is 4.6 * 10^6 J.
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
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