q1 = 1000g x 2.097 kJ/gK x (100.0 - 120.0) = -209.7 kJ
q2 = 1000g x 2257 kJ/kg = 2257 kJ
q3 = 1000g x 4.184 kJ/gK x (90.0 - 100.0) = -418.4 kJ
Total heat removed = q1 + q2 + q3 = 2257 - 209.7 - 418.4 = 1629.3 kJ/kg
Spent steam from an electric generating plant leaves the turbines at 120.0 degrees Celsius and is cooled to 90.0 degrees Celsuius liquid water by water from a cooling tower in a heat exchanger. How much heat is removed by the cooling tower water for each kilogram of spent steam?
q1 = heat removed by moving steam from 120.0 degrees C to steam @ 100.0 degrees C.
q1 = mass x specific steam steam x (Tf-Ti).
q2 = heat removed by condensing steam @ 100.0 degrees C to liquid water at 100.0 degrees C.
q2 = mass x heat of condensation (same as heat of vaporization but different sign).
q3 = heat removed by moving liquid water from 100.0 degrees C to 90.0 degrees C.
Total heat removed is q1 + q2 + q3.
The problem doesn't exactly specify a mass but it asks for heat per kg so I would use 1 kg or 1000g for mass in each of the above equations.
1 answer
1 answer