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Calculate the change in heat when 23.95 grams of water vapor (steam) at 100 degrees celsius condenses...?
Calculate the change in heat when 23.95 grams of water vapor (steam) at 100 degrees celsius condenses to liquid water and then cools to 1.00 degrees celsius. =________ joules
Thank you for any help you can give me I appreciate it! I have been shown it once on a similar kind of problem, but can't get it for this one.
1 Answer
- ?Lv 710 years agoFavorite Answer
Latent Heat of Vaporisation/Condensation is 2,260J/g.
Specific heat of water is 4.184 J/g/°C.
1....To condense the steam.
mcΔT = Q .
23.95g x 2,260J/g = -54,127 (Joules of heat released).
2...To cool the water to 1.0°C
mcΔT = Q.
23.95g x 4.184J/g/°C x 99°C ΔT = -9920.5 (Joules of heat released).
Total heat released = -54,127J + -9,920.5L = -64,047.5 Joules of heat released
