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.
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.
-
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
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