6,520 Btus
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
Assuming standard atmospheric pressure, 2260 kilojoules.
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.
Raise the temp of 52 grams of water from 33.0 C to 100 C = 52*67*4.184 = 14.577 kJConvert evaporate 52 g of water to steam without change of temp = 52*2259 = 117.468 kJRaise the temp of 52 grams of steam from 100 C to 110 C = 52*10*2.02 = 1.051 kJTotal energy required = 133.095 kJ = 31,811 calories or 31.811 kCal.
2000 lbs
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.
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