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How much energy will be released when 25g of steam at 100 degree Celsius converts into liquid water at 100 degree Celsius?
Wiki User
∙ 9y ago
This is fairly easy to do.
Q = M.Hevap
Q = energy for phase change
M = mass
HEvap = heat of vaporization
The heat of vaporization can be obtained from steam tables, usually with a unit of kJ/kg. Convert your mass to kg and then find Q (kJ).
Then check for a conversion factor between kJ and calories.
Wiki User
∙ 14y ago
Wiki User
∙ 11y agoIf the water is at 100 degrees Celsius, then it has already reached boiling point.
If you meant to ask how much energy is needed to vaporize (turn from liquid to gas) the water, do the following:
1. convert grams to moles of water
45.00 grams H20 x mol/18.016 grams = 2.50 mol H2O
2. find needed kJ using the heat of vaporization conversion factor.
2.50 mol x 40.6 kJ/mol H20 = 102 kJ are needed to convert 45.0 grams of liquid water to vapor.
Wiki User
∙ 9y agoThe heat released, QR , is given by the following :
QR = ( m ) ( h sub fg )
where h sub fg is the latent enthalpy of vaporization of water at 100.0 C.
QR = ( 25 g ) ( 2256.4 J/g ) = 56410 J = 56.41 kJ <---------
The value of h sub fg was obtained from Steam Table at 100.0 C.
Wiki User
∙ 13y agoIts 45,849 cal , but round to significant fig. to 46000 cal. and u just had to multiply the mass x the heat of vaporization that is given to you, and that's it!!!!
Wiki User
∙ 11y ago540 + (100 -22) = 618 cal
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