The specific heat capacity of water between 25 C and 100 C is 4.1813 J / (g*K).
Beyond 100 C, the heat capacity of water is 2.080 J / (g*K)
So, it take 4.1813 joules of energy to heat 1 gram of water 1 degree Celsius (Kelvin).
Therefore, in order to heat 22 g of water from 25 C to 100 C (75 C), it would require:
4.1813 * 22 * 75 = 6899.145 J
And, to heat 22 g of steam from 100 C to 125 C (25 C), it would require:
2.080 * 22 * 25 = 1144 J
The combined amount of energy required would be:
6899.145 + 1144 = 8043.145 J
at 100 degrees liquid water will go to steam and steam will go to liquid water
As the steam comes in contact with the skin, it becomes water, and releases more energy (about 2188 joules per gram) on contact than water at the same temperature.
1 BTU is the energy required to heat 1 pound of water by 1 degree F. 1 Joule is defined mechanically, but in thermal terms it is 1/4.2 of a calorie (4.2 Joules/calorie), and 1 calorie is the energy required to heat 1 gram of water by 1 degree C. In fact 1 BTU = 1055 Joules.
Water is vapor or steam at 213 0F at normal pressures.
It's necessary to remove 540 calories from a gram of water in order to freeze it. That's about 2260 joules. The amount of energy used by a freezer to do this depends on the efficiency of the freezer.
15480.80
The latent heat of condensation of steam is 2260 Joules per gram (539.3 cals/g). So the amount of heat released by 12.4 g = 12.4*2260 Joules = 28,024 Joules or 6687 cals.
Can you help
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.
2830 g of water raised through 50 degrees C would use 2830 x 50 calories. But then to boil the water away to steam completely requires another 550 calories per gram, which is 2830 x 550 calories. To convert to Joules, use 4.2 Joules per calorie.
Heat of vaporization of water is 2.26 x 106 joules per kg. Therefore 1 gram of water will need 2.26 x 103 joules.
At standard pressure a gram of water at 100 degrees C requires 550 calories to convert it into steam at 100 degrees C. Therefore, 234.5 grams requires 128975 calories, or 541700 Joules.
A watt is a unit of power: what is required is probably the amount of energy - which is measured in joules.
334.8 Joules
It equals one kilpod.
70 calories per gram. (The specific heat capacity of water is 1 calorie per gram per degree C.) This could be converted into Joules if necessary using the conversion factor of 1 calorie = 4.18400 Joules.
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.