Donanderson
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
Wiki User
∙ 13y ago6,520 Btus
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
6,520 Btus
To change 10 pounds of ice at 20 degrees Fahrenheit to steam at 220 degrees Fahrenheit, you need to supply enough energy to first melt the ice, then heat the water to the boiling point, and finally convert it to steam. This process requires approximately 180 BTUs per pound of ice to melt it, 180 BTUs per pound of water to heat it to the boiling point, and then 970 BTUs per pound of water to convert it to steam. So, for 10 pounds of ice, the total BTUs required would be around 18,300 BTUs.
To change 5 pounds of ice at 20°F to steam at 220°F, you will need to go through multiple phases: raise ice temperature to 32°F, melt ice to water at 32°F, raise water temperature to 212°F, and then convert water to steam at 212°F to steam at 220°F. The total heat required, in BTUs, is around 503 BTUs per pound of ice, which translates to about 2515 BTUs for 5 pounds of ice.
To calculate the heat energy required to change the temperature of a substance, you can use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Given that steam has a specific heat capacity of approximately 0.5 btu/lb°F, you can calculate the required heat energy by plugging in the values: Q = 10 lb * 0.5 btu/lb°F * (240°F - 212°F).
To change 1 gram of ice at 22 degrees Fahrenheit to steam at 212 degrees Fahrenheit, you would need approximately 1064.73 BTUs. This calculation takes into account the energy required to melt the ice, heat the water, and then boil it to steam.
it takes 2 pounds of it
To calculate the heat required to cool steam to water at a lower temperature, you can use the formula: Q = mcΔT Where: Q = heat energy m = mass c = specific heat capacity ΔT = change in temperature Given: m = 5 pounds ΔT = (232 - 162) = 70 degrees F Specific heat capacity of water = 1 cal/g°C Convert pounds to grams: 1 pound ≈ 453.592 grams Now, plug in the values into the formula to calculate the heat energy.
Water changes into steam or water vapor above 100 degrees Celsius.
It equals one kilpod.
To change 5 pounds of ice at 20°F to steam at 220°F, you would first need to heat the ice to its melting point, then heat the water to its boiling point, and finally convert the water to steam. The total heat required can be calculated using the specific heat capacities of ice, water, and steam, as well as the heat of fusion and vaporization. The specific calculations would depend on the specific heat capacities and heat of fusion/vaporization values provided.
Assuming standard atmospheric pressure, 2260 kilojoules.
Steam has more heat energy compared to water at 100 degrees Celsius because steam is in a gaseous state and has absorbed additional heat energy to transition from a liquid to a gas.