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That completely depends on how much steam there is. (mass)
1,000 joules of heat energy for every second that the 1,000 watts' dissipation continues.
The necessary heat is 9,22 joules.
300 calories, which is equal to about 1255.2 joules
46389000 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.
That completely depends on how much steam there is. (mass)
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.
If all 1700 Joules of work get converted into heat, then, of course, you get 1700 Joules of heat.
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.
It is measured in joules (J)
About 322.5 Joules of heat
1,000 joules of heat energy for every second that the 1,000 watts' dissipation continues.
Assuming atmospheric pressure, ice of 0oC and steam of 100oC? This equals the heat of fusion + the specific heat of water * 100 + the heat of vaporization At 1 bar (rougly 1 atm): heat of fusion = 333.55 j/g specific heat = 4.186 j/(g*k) * 100 = 418.6 j/g heat of vaporization = 2257 j/g So 333.55 + 418.6 + 2257 = 3009.15 joules required.
0.131 joules/gram'C x 1.3 grams x (46-25)'C = 3.5763 joules
The latent heat for water to steam is 550 calories (2310 Joules) per gram. For 2 grammes, double these figures.
5 joules