100
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
38.2C-24.5C = 13.7 C change this requires 13.7 calories per gram or 1370 calories total
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
To raise the temperature of one cc of water requires i calorie of heat . you did not specify the volume.
1370 calories
there are 3000 calories in a water bottle so be careful!
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
(5)(3)= 15 calories. 1 calorie is the energy (heat) to raise 1 gram of water by 1 degree celsius, so 5 grams of water (3 degrees Celsius) = 15.
38.2C-24.5C = 13.7 C change this requires 13.7 calories per gram or 1370 calories total
A calorie is the amount of heat you need to raise the temperature of one gram of water by one degree Celsius. Assuming you are raising the temperature of the water from twenty degrees Celsius to ninety-nine degrees Celsius, it would take 20,000 calories. To calculate this, subtract 20 from 99. This is the amount of degrees you need to raise the temperature of the water by. Then multiply that number by 256, the amount of water in grams. You should get 20,244 calories. In significant digits, your answer should be 20,000 calories.
21 grams through 71 degrees is 21x71 calories.
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
103 calories. The heat of fusion of water is 80 cal/g and it takes one calorie to change the temperature of 1 g of water by 1 degree. 80+23=103 calories
I don't think you actually put calories into the water. You would dissipate thermal energy that could be measured in kCal or calories to heat the water. You may be thinking of the stored thermal energy from the heated water.
This heat is 51, 33 cal.