Spent steam from an electric generating plant leaves the turbines at 120.0 degrees celsius and is cooled to 90 degrees celsius liquid water by water from a cooling tower in a heat exchanger. How much heat is removed by the cooling towerwater for each kg of spent steam?
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1 answer

to get the heat released or absorbed,
Q = mc(T2T1)
where
m = mass of substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
**note: if Q is (), heat is released and if (+), heat is absorbed
now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from vapor>liquid
thus we need another data called Latent Heat of Vaporization to calculate for the heat required to change its phase:
H = m(Lv)
where
m = mass
Lv = Latent Heat of Vaporization
**note: H is () if heat is released and (+) if heat is absorbed
Q, total = Q1 + H + Q2
Q,total = mc(T2T1) + mLv + mc(T3T2)
where
T3 = 90 deg
T2 = 100 deg (boiling point of water)
T1 = 120 deg
since heat is removed, H is negative. now for every 1 kg of steam,
Q,total = 1*(c)*(100120)  1*Lv + 1*(c)*(90100)
Q,total = 20*c  10*c  Lv
Q,total = 30*c  Lv
now you look for the specific heat of water, c, and latent heat of vaporization, Lv. note that units must already be in Joules or calories.
hope this helps~ :) 👍
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answered by Jai
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