All calculations below are assumed done with all materials (ice, water, and steam) being at normal atmospheric pressure.
3.0kg is 3000 gm. Heat required to raise temperature of ice from -10 degree celcius to 0 degrees celcius is 3000 x 10 cals. (A)
The latent heat of fusion of ice is 80 cals per gm. So heat needed to melt 3000 gm of ice at 0 degrees celcius is 3000 x 80 cal. (B)
The heat required to convert 3000 gms of water from 0 degrees celcius to 100 degrees celcius is 3000 x 100 cals. (C)
The latent heat of vaporization of steam is approximately 540 cals per gm. Hence heat required to convert 3000 gms of water to steam at 100 degrees celcius is 3000 x 540 cals. (D)
Now add (A) + (B) + (C) + (D) = 30000 + 240000 + 300000 + 1620000 cals.
And that is = 2190000 cals = 2.19 X 10 to power 6 cals.
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
For one gram of ice, it takes 11.9 calories to change the temperature to 0°, 80 calories to melt the ice, 100 calories to raise the water temperature to 100°, 540 calories to change the water to steam, and 23 calories to raise the steam temperature to 123°. That's a total of (11.9 + 80 + 100 + 540 + 23) calories or 754.9 calories. So to do the same to 55.6 grams of ice requires 55.6 times as much heat. 754.9 calories times 55.6 equals approximately 41972 calories (about 42 kilocalories).
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.
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.
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.
4.2 × 105 J
100 degree Celsius
Quantity of Energy= mass x temperature change x specific heat capacity For example: Find the amount of energy needed to raise the temperature of 0.20 kg of lead by 15 degree Celsius if the specific heat capacity of lead is 0.90 J/g degree Celsius. Answer: J=200g x 15 degree Celsius x 0.90 J/g degree Celsius = 2700 J
Approx 2940 Joules.
Steam. Water boils at 100 degrees Celsius.
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
It takes 80 calories per gram to increase the temperature of water by one degree. 4000 * 80 * 100 = 32000000 calories.
540 cal
heat energy required to raise the temperature of ice by 29 celsius =specific heat capacity of ice * temperature change *mass of ice + to change 1kg of ice at 0 celsius to water at 0 celsius =specific latent of fusion of ice*mass of water + heat energy required to raise the temperature of water by 106 celsius =specific heat capacity of water * temperature change *mass of ice + to change 1kg of water at 106 celsius to steam at 106 celsius =specific latent of fusion of ice*mass of steam
steam. It has to go through a phase change, which takes additional energy to get there.
Between 45 and 50 Degree Celsius, depending on your body and the ability for it to handle the steam. Breathing becomes harder and heart rate goes up!
Water or steam at 100 degrees Celsius will kill some bacteria. Detergent or soap is needed to kill the ones that survive the hot temperatures.