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?

1 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.

   Answer ID
   539640

   Created
   April 30, 2011 7:30pm UTC

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   https://questions.llc/answers/539640