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.
Leaving aside the effects of pressure, yes, the temperatures are the same. But the amount of heat (thermal energy) per gram, is much greater for the steam.
Assuming standard atmospheric pressure, 2260 kilojoules.
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.
Raise the temp of 52 grams of water from 33.0 C to 100 C = 52*67*4.184 = 14.577 kJConvert evaporate 52 g of water to steam without change of temp = 52*2259 = 117.468 kJRaise the temp of 52 grams of steam from 100 C to 110 C = 52*10*2.02 = 1.051 kJTotal energy required = 133.095 kJ = 31,811 calories or 31.811 kCal.
When steam at 100 degrees Celsius condenses, it releases 2260 Joules of energy per gram. Therefore, for 1000g of steam, the heat released would be 2,260,000 Joules (2260 J/g * 1000 g).
Steam at 100 degrees Celsius is more dangerous than water at 100 degrees Celsius because steam carries more heat energy and can cause deeper burns on contact with skin. Additionally, steam has a greater ability to penetrate clothing and protective gear, increasing the risk of burns.
at 100 degrees liquid water will go to steam and steam will go to liquid water
When water reaches boiling, it becomes steam. Once the water cools down, typically below 100 degrees Fahrenheit, it will begin to form water droplets or condensation.
The most probable answer for your your question is pure water.
The heat released when steam at 100 degrees Celsius condenses into liquid water is equal to the heat it absorbed during the phase change from liquid water to steam. This heat is known as the latent heat of vaporization, which for water is approximately 2260 J/g. Therefore, the amount of heat released when 160 grams of steam liquefies is approximately 160g x 2260 J/g = 361,600 J.
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.
Water becomes steam at 100 degrees Celsius under normal atmospheric pressure.
If you heat steam above 100 degrees Celsius, it will continue to increase in temperature without changing to water. This is because steam is already in a gaseous state at temperatures above 100 degrees Celsius.
It can be anywhere in between 60C and 100C (boiling point).
Yes, it can.
To convert 12.5 grams of ice at 0 degrees Celsius to steam at 100 degrees Celsius, you would need to provide heat energy for three main processes: heating the ice from 0 degrees Celsius to 100 degrees Celsius, melting the ice into water at 0 degrees Celsius, and then heating the water from 0 degrees Celsius to steam at 100 degrees Celsius. The total calorie requirement would be determined by the specific heat capacities and heat of fusion and vaporization of water.