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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. to get the heat released or absorbed,
    Q = mc(T2-T1)
    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(T2-T1) + mLv + mc(T3-T2)
    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)*(100-120) - 1*Lv + 1*(c)*(90-100)
    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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