212-65=147. 147(20)=2940 btu needed is this right?
25
300
12oC
100 calories. 1 calorie is defined as the amount of energy required to raise the temperature of 1 gram of 1 by 1 degree Celsius. So, if you need to raise 10 grams of water 1 degree, you would need 10 calories of energy. If you needed to raise those same 10 gram by 10 degrees, you'll need 10 * 10, or 100 calories.
A quart of water is 946.35 cubic centimeters and since density of water is 1 gram per cubic centimeter the weight of a quart of water is 946.35 grams. Heat required = (mass) x (specific heat of substance) x (temperature differential) In our case it would be 946.35 x 1 x 38(assumed degrees centigrade) = 35961.3 cals
2250
25
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.
To calculate the BTUs required to raise the temperature of 15 pounds of water, you can use the formula: BTUs = Weight of water in pounds × Temperature change in degrees Fahrenheit × 1 BTU So, the calculation would be: BTUs = 15 lbs × (130°F - 100°F) × 1 BTU = 15 lbs × 30°F = 450 BTUs.
10 x (70 - 50) = 200 Btu
200 BTU. I'm assuming your temperatures are in Fahrenheit, since all of your other measurements are in the Imperial system. A BTU is defined as the amount of energy required to raise the temperature of 1 pound water by 1 degree F, so the temperature is raised by 20 degrees, and 10 pounds of water: 20 x 10 = 200
U.S.gallon = 8.33 pounds of water. Therefore to raise the temperature by one degree F will require 8.33 BTU. The initial temperature of 50 F is inconsequential.
That will completely depend on how much water there is.
To convert 7975 pounds of water weight into degrees, you need to know the temperature in which the water is being measured. Degrees typically refer to temperature, not weight, so more information is needed to provide an accurate conversion.
It takes 1 calorie to raise the temperature of 1 gram of water by 1 degree Celsius. Therefore, to raise the temperature of 1 gram of water by 5 degrees Celsius, it would require 5 calories.
It takes 6 SI calories to raise one liter of water by 6 degrees Celsius.
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