What is the total energy transfer(in KJ) when 10.00 pounds of water vapor at 110 degrees C becomes ice at 253K? Assume that the specific gravity of water = 1.00. The answer I got was around 12,000 KJ. Is this correct?

Answer 1

10 pounds of water? Goodness.

10 lbs *1kg/2.2lbs=you do it.

Then,
Heatcoolingsteamto100C=masswater*specificheatsteam*(100-110)
heat condensing steam= masssteam*HvSteam
Heatreleasedcoolingwater= masswater*specificheatwater*(0-100)

all those should be negative, add them,that is the heat gained total, so the heat released willbe the negative of that.