The amount of energy that is required to 160 cfm of air from 10 to 170 degrees F is 200 btu. T he formula is weight x specific heat x temperature difference so we have10 pounds x 1.00 x 2010 for 10 pounds of water.
The energy required to raise the temperature 1 degree Celsius of 1 gram of water (1 mL) is 1 calorie (=4.18 J). So for 1 kg, 1Kcal (= 4180 J = 4.18 KJ) is required. To raise it 60 degrees, just multiply by 60 and for 10 kg multiply by 10 again. That would make 2.508 MJ (= 2508000 J) Now this is not completely accurate. The energy required to raise the temperature of water differs at 20 degrees from that at 60 degrees. The difference is small (~0.05 J or something like that) but still present.
538J
That will completely depend on how much water there is.
That's going to depend on how much water you're responsible for. Teacup at 60 degrees . . . very few BTU. Swimming pool at 60 degrees . . . many more BTU. It's also going to depend on whether you're talking about Celsius or Fahrenheit degrees. Fahrenheit degrees . . . fewer BTU. Celsius degrees . . . more BTU. (Also, the water will escape as you pass 100.) In general, one BTU is approximately the energy required to raise the temperature of 1 pound of water 1 degree Fahrenheit. You can take it from there, when you reach the job site and determine the exact scope of the work.
It doesn't work that way. There is not a certain number of btus to raise air temperature. You would have to know how much air. A BTU is the British Thermal Unit. That is the amount of energy required to raise the temperature of one pound of water one degree F.
13,455 J
314j
3.50 J
1935 J (apex)
1935 J (apex)
4.1858 joules of energy will raise the temperature of 1 g of water by 1oC. Thus, 4.1858 * 955 * 80 = 319795.12 joules of energy is required to raise the temperature of 955 g of water by 1oC.
The energy required to raise the temperature 1 degree Celsius of 1 gram of water (1 mL) is 1 calorie (=4.18 J). So for 1 kg, 1Kcal (= 4180 J = 4.18 KJ) is required. To raise it 60 degrees, just multiply by 60 and for 10 kg multiply by 10 again. That would make 2.508 MJ (= 2508000 J) Now this is not completely accurate. The energy required to raise the temperature of water differs at 20 degrees from that at 60 degrees. The difference is small (~0.05 J or something like that) but still present.
How much heat energy is required to raise the temperature of 0.358 of copper from 23.0 to 60.0 ? The specific heat of copper is 0.0920
21 Kg = 2100 grams to rise the temperature of this amount of water by 2 degrees Celsius you need 2*2100 = 4200 calories or 17572.8 Joules.
The needed energy is 10 calories.
It depends on the specific energy of the substance.
1935