Want this question answered?
E = m c (delta)T
steam. It has to go through a phase change, which takes additional energy to get there.
Water turns to steam (or in other words, it boils) at 100 degrees Celsius or 212 degrees Fahrenheit.
Steam is produced whenever water boils, which occurs at 100 degrees Celsius at standard pressure and temperature (boiling point changes at altitudes because of the pressure change, remember).
heat energy required to raise the temperature of ice by 29 celsius =specific heat capacity of ice * temperature change *mass of ice + to change 1kg of ice at 0 celsius to water at 0 celsius =specific latent of fusion of ice*mass of water + heat energy required to raise the temperature of water by 106 celsius =specific heat capacity of water * temperature change *mass of ice + to change 1kg of water at 106 celsius to steam at 106 celsius =specific latent of fusion of ice*mass of steam
6,520 Btus
E = m c (delta)T
1 BTU is required to raise 1lb of water 1 degree F in 1 hour. 212-75=137 degrees 600 lbs water x 137 degrees= 82,200 BTU's required to change 75 degree water to 212 degree water. To change 212 degree water to 212 degree steam it requires 970 btu's (latent heat of vaporization) per lb of water 970 btu x 600 lbs water = 582,000 btu Answer - 582,000 btu+ 82,200 btu = 664,200 btu's
The question cannot be answered because:the temperature scale being used has not been specified,There is no normal temperature scale in which you can have ice at 32 degrees and steam at 82 degrees without large changes in pressure. If changes in pressure are permitted then there is no simple formula to calculate the amount of heat (btus) required.
it takes 2 pounds of it
Steam. Water boils at 100 degrees Celsius.
steam. It has to go through a phase change, which takes additional energy to get there.
It equals one kilpod.
46389000 j
Assuming standard atmospheric pressure, 2260 kilojoules.
Water turns to steam (or in other words, it boils) at 100 degrees Celsius or 212 degrees Fahrenheit.
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.