If 144 kJ of Energy were removed from 55.0 g of water vapor at 100 degrees Celcius, what would be the final temp of the water?
Heat of Vaporization/Condensing = 2260 J/g
Heat Capacity of Liquid Water= 4.18 J/ g*C
Q=m c (delta temp change)
- Roger the MoleLv 72 years agoFavorite Answer
(2260 J/g) x (55.0 g) = 124300 J = 124.3 kJ for condensing the steam
(144 kJ total) - (124.3 kJ) = 19.7 kJ remaining to cool the condensed steam
(19.7 x 10^3 J) / (4.18 J/g·°C) / (55.0 g) = 85.7°C change
100°C - 85.7°C = 14.3°C
- 2 years ago
I got 14.3 degrees Celsius as my answer but I’m not very confident.