PHYSICS help please!!!!!!!!!!

At a nuclear power plant water at 8°c from a lake is used in a heat exchanger to condense spent steam at a temperature of 120°c,to water at 85°c before it is recycled to the reactor. If the cooling water returns to the lake at a temperature of 19°c,how many kilogram of water are needed per kilogram of steam ? Ignore the pressure change in steam.

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asked by Riana
  1. Q(gained) =Q(lost)
    Q(gained) =Q₁+Q₂+Q₃
    Q₁ = c₁m(Δt₁)=2020•1•(120-100) = 40400 J
    Q₂ = Lm=2256000•1= 2256000 J
    Q₃=c₂m(Δt₂)=4186•1•(100-85) = 62790 J
    Q(gained) =40400 +2256000+62790 =2359190 J
    Q(lost) =c₂•M•(Δt₃)=4186•M•(19-8) = 4186•11•M
    M= Q(gained)/ 4186•11=2359190/4186•11= 51.2 kg

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    posted by Elena
  2. Can you please explain it to me...that what's happening in all the process?

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  3. Q(gained) =Q(lost)
    Q(lost) =Q₁+Q₂+Q₃
    Q₁ = c₁m(Δt₁)=2020•1•(120-100) = 40400 J
    Q₂ = Lm=2256000•1= 2256000 J
    Q₃=c₂m(Δt₂)=4186•1•(100-85) = 62790 J
    Q(lost) =40400 +2256000+62790 =2359190 J
    Q(gained) =c₂•M•(Δt₃)=4186•M•(19-8) = 4186•11•M
    M= Q(lost)/ 4186•11=2359190/4186•11= 51.2 kg
    The processes in which there is a loss of energy :
    steam cooling up to 100⁰ (Q₁), ‘steam-water’ transformation (Q₂), and water cooling up to 85⁰ (Q₃)
    This energy heat the water from the lake from 8⁰ to 19⁰ (Q(gained))

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    posted by Elena

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