756.78 degrees Celsius
4.2 × 105 J
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
As a rule of thumb you would need about 138 Kg of coal (26GJ/Ton) to produce 1 ton of steam.
63 kJ is needed.
it depends how cold the ice is
i think 1 kg steam
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)
4.2 × 105 J
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
46389000 j
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.
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