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
The heat lost by 1 gram of water at 0 degrees Celsius as it freezes to form ice is approximately 333.55 joules. This is the heat of fusion of water, which is the energy required to change the state of water from a liquid to a solid at its melting point.
At 105 degrees Celsius, steam will remain in the gaseous phase as it is above the boiling point of water (100 degrees Celsius). Steam will continue to condense into liquid water only once it cools down below the boiling point.
Steam at 100 degrees Celsius contains more heat energy compared to liquid water at the same temperature. When steam comes into contact with skin, it releases this extra heat quickly, causing more severe burns compared to liquid water of the same temperature.
Water produces steam at a temperature of 212 degrees Fahrenheit (100 degrees Celsius) at sea level. This is known as the boiling point of water. When water reaches this temperature, it changes from a liquid state to a gaseous state, forming steam.
To bring the ice block to 0 degrees Celsius, you would need 150,000 Joules (Q = mcΔT). To melt the ice at 0 degrees Celsius, you would need 3,375,000 Joules (Q = mLf). Heating the water from 0 to 100 degrees Celsius would require 1,500,000 Joules (Q = mcΔT). Turning the water to steam at 100 degrees Celsius would need 10,500,000 Joules (Q = mLv). Finally, heating the steam to 120 degrees Celsius would require 600,000 Joules (Q = mcΔT). In total, you would need 15,125,000 Joules of heat energy.
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.
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To heat 1 gram of water by 1 degree Celsius, it takes 4.18 joules. So, to heat water from, for example, 20 degrees to 100 degrees, you would need to calculate the total mass of water and apply the specific heat capacity to determine the total energy required.
To calculate the energy released when the steam cools to water, you need to consider the specific heat capacity of water and steam. The equation Q = mcΔT can be used, where Q is the energy released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Once you have the energy released, you can convert it to joules.
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 100 degrees liquid water will go to steam and steam will go to liquid water
The specific heat capacity of water is 4.18 J/g°C. To calculate the energy required to raise 21 kg of water by 2 degrees Celsius, use the formula: Energy = mass x specific heat capacity x temperature change. Plugging in the values, the energy required is 21,084 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.