To calculate the BTUs needed to convert one pound of water at 200°F to ice at 30°F, we must consider two stages: first, cooling the water from 200°F to 32°F, and then freezing it to ice at 32°F, and finally cooling the ice to 30°F.
Adding these together: 168 + 80 + 2 = 250 BTUs must be removed.
313 Btu
LATENT HEAT OF FUSION When one pound of ice melts, it absorbs 144 BTUs at a constant temperature of 32°F. If one pound of water is to be frozen into ice, 144 BTUs must be removed from the water at a constant temperature of 32°F.
One pound of steam at 212 degrees Fahrenheit contains 1150 BTUs of energy.
144
It takes 180 BTUs to raise 1 pound of water from 32 degrees Fahrenheit to 212 degrees Fahrenheit to convert it completely into steam. This change in temperature includes heating the water from its freezing point to boiling point, then undergoing phase change from liquid to gas.
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 raise the temperature of 1 pound of water by 1 degree Fahrenheit, it requires 1 BTU (British Thermal Unit). Therefore, to heat water by 10 degrees, it would require 10 BTUs per pound of water. This value may vary slightly depending on the specific heat capacity of the water.
To determine the BTUs that must be removed from one pound of water at 200°F to convert it to ice at 30°F, we need to account for several steps: cooling the water from 200°F to 32°F (the freezing point), the phase change from water to ice at 32°F, and then cooling the ice from 32°F to 30°F. Cooling the water from 200°F to 32°F requires about 168 BTUs (1 BTU cools 1 pound of water by 1°F). Freezing the water at 32°F requires the removal of 80 BTUs (latent heat of fusion). Cooling the ice from 32°F to 30°F requires an additional 2 BTUs. In total, approximately 250 BTUs must be removed (168 + 80 + 2 = 250 BTUs).
One BTU is the energy required to raise one pound of water by one degrees. Therefore, your answer would be one half.
actually its 313.
To raise 1 pound of ice from 32°F to water at 32°F it requires 144 BTUs. Since you have 50 pounds of ice, you would need 50 * 144 BTUs to raise the ice to water at 32°F. To further raise the water from 32°F to 160°F, you would need an additional amount of BTUs based on the specific heat capacity of water.
212 - 80 = 132 degrees temperature increase x 1 pound water = 132 BTU