756.78 degrees Celsius
The amount of heat that must be removed from steam to change it to a liquid is the latent heat of vaporization of water, which is approximately 2260 kJ/kg at atmospheric pressure. This is the amount of energy required for water to change from a gas to a liquid state at a constant temperature.
It depends on the temperature of the water that you begin with. But lets say that you start with 100 Kg of water at 100'C. the heat of vapourization of water is 2260 J /g or 2,260,000 J/Kg Since we have 100Kg it would require 226,000,000 J converting Joules into Kilowatt hours gives 226 MegaJoules as 1 kilowatt hour is equal to 3.6 megajoules 226/3.6 =62.77 kilowatt hours
The process involves increasing the temperature of water from 8°C to 100°C and then changing its phase to steam at 100°C. The total heat energy required can be calculated using the specific heat capacity of water and the heat of vaporization. The formula Q = mcΔT can be used to find the heat energy needed, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change.
The heat required to melt 5 kg of ice is 334,750 J. This is calculated using the formula Q = m * Lf, where Q is the heat energy (Joules), m is the mass of the substance (kg), and Lf is the specific latent heat of fusion for ice (334,750 J/kg).
The heat of vaporization of mercury is 59.11 kJ/kg. To convert this to joules, we multiply by 1000, which gives us 59,110 J/kg. Therefore, the energy released when 0.06 kg of mercury is condensed to a liquid at the same temperature would be 0.06 kg * 59,110 J/kg = 3,546.6 J.
1 kg of steam at 373 K contains more heat than 1 kg of water at 373 K because steam has a higher specific heat capacity and latent heat of vaporization than water. This means more heat energy is required to convert water at 373 K into steam at 373 K.
when steam condenses it gives out large quantity of latent heat(2260000 J/kg).thus the burns caused by steams affect a larger and deeper area.
Heat required to have such a change of state is called latent heat. If L J/kg is the latent heat per kg of water then for M kg of water we need M* L joule of heat energy
Latent heat of evaporation of water to steam is 2270 KJ/Kg
The difference is the evaporation heat (or the 'equal' condensation heat)
for converting cubic meter to ton , density or specific volume is needed. specific volume unite is m3/kg. steam cubic meter/ (cubic meter/kg)= steam (kg ) /1000= tone of steam
The amount of heat that must be removed from steam to change it to a liquid is the latent heat of vaporization of water, which is approximately 2260 kJ/kg at atmospheric pressure. This is the amount of energy required for water to change from a gas to a liquid state at a constant temperature.
You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.
Assuming standard atmospheric pressure, 2260 kilojoules.
You might take a known quantity of water, add heat and stir until all of it has reached a temperature of 100 C (212 F), then start measuring how much more heat it takes to turn all of the water to steam. That's not exactly a 'derivation'. It's more like an empirical measurement, and we're thinking that's how the figure was originally determined.
2260 kj/kg X 0.086 kg = 194 kj The heat of vaporization for water is 2260 kj/kg at 1 atmosphere pressure.
Question is wrong