Approx 2940 Joules.
539 calories per gram for heat of vaporization plus 1 cal/gram/degree C 100 degrees C - 80 degrees C = 20 degrees C (539 calories + 20 calories) X 50 kg X 1000 gm/kg = 27950000 cal = 27,950 kcal
That completely depends on how much steam there is. (mass)
1
This value is 22,418 kJ.
steam scalds are more serious than boiling water scalds because steam when condenses onto the more cooler skin, it loses latent heat of vaporization (to become water at 100 degrees Celsius) also it loses thermal capacity to become equal to the temperature of the skin (37 degrees Celsius). boiling water loses only thermal capacity as it cools down to 37 degrees Celsius from 100 degree Celsius.
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.
Use this formula. q(in Joules) = Mass * specific heat * change in temperature I will use specific heat of water at 25 C. You can look up specific heat of steam. You say " heat to " so I assume you have final and initial heat backwards. q = 25.0 grams H2O * 4.180 J/gC * (100.0 C - 29.25 C ) q = 7393 Joules
4.2 × 105 J
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.
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.
steam. It has to go through a phase change, which takes additional energy to get there.
539 calories per gram for heat of vaporization plus 1 cal/gram/degree C 100 degrees C - 80 degrees C = 20 degrees C (539 calories + 20 calories) X 50 kg X 1000 gm/kg = 27950000 cal = 27,950 kcal
That completely depends on how much steam there is. (mass)
1
A 300 grams of water takes about 90 seconds to boil in a 2 kW kettle, so that is 2000 watts x 90 seconds which is 180,000 Joules of energy. But to convert that water completely into steam requires an extra 300x550x4.2 Joules, which is nearly 700,000 Joules. So converting it to steam takes 4-5 times as much energy as boiling it. That is why it takes a while for a kettle to boil dry.
This value is 22,418 kJ.
E = m c (delta)T