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)

Q=m (Hv)

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  • 2 years ago
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    (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

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  • 2 years ago

    I got 14.3 degrees Celsius as my answer but I’m not very confident.

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