PHYSICS PLEASE HELP DUE SOON

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Carnot engine operates between 170°C and 23°C. How much ice can the engine melt from its exhaust after it has done 4.5 104 J of work?

  • PHYSICS PLEASE HELP DUE SOON -

    Post what you have so far, and I can guide you.

  • PHYSICS PLEASE HELP DUE SOON -

    T1=273+170=443 K
    T2=273+23=296 K
    efficiency =(Q1-Q2)/Q1 =W/Q1
    efficiency =(T1-T2)/T1
    W/Q1 = (T1-T2)/T1
    Q1=W•T1/(T1-T2)
    Q1= r•m
    r= 333•10⁵ J/kg
    W•T1/(T1-T2) =r•m
    m = W•T1/(T1-T2)• r =
    (4.5x10^4)•443/(443-296)•333•10⁵=4.5•10^12

  • PHYSICS PLEASE HELP DUE SOON -

    A neat trick:

    vbr{lr<WT1x^(1/9)><f(Q1Q2)/.bbn>
    vbr<lr><5.12>(296^3)
    .bbn=%eff^5
    %ff=F(443)
    .jrbn=%44x10^2
    .qen=.133%tri
    .yyn=.tri[kg]
    t.r(333x10^5)i=vBr2^(1/3)
    eff1=31.4%eff2
    t4=59.3J/kg
    t2=28.6J/kg

    Tell me what you get:

  • PHYSICS PLEASE HELP DUE SOON -

    I am really not sure how to do this

  • PHYSICS PLEASE HELP DUE SOON -

    I don't understand the abbreviations and symbols { <>, .yyn .jrbn etc

  • PHYSICS PLEASE HELP DUE SOON -

    What I did:
    1)vbr function--easiest to use
    2).bbn is to solve .bbn is easy to the %eff^5
    3)Set %ff to F(T1)
    4)Make function of .jrbn
    .jrbn=%continual change or %eff^5 which came out to 44x10^2
    5)Function of .qen= 133% of tri or bbn final.
    6)Function of .yyn=tri what you just solved for.
    7)Rearrange tri which is equal to 1.33
    1.3(333x10^5)i = vBr2^(1/3)
    Solved for % of eff3
    8)eff1=31.4%eff2--self explanatory
    9)then you get t4 and t2
    10)Solve from that

  • PHYSICS PLEASE HELP DUE SOON -

    If I didn't elaborate enough on step 7 you solve for vBr2^(1/3) for eff3

  • PHYSICS PLEASE HELP DUE SOON -

    I get 2193.17

  • PHYSICS PLEASE HELP DUE SOON -

    I give up

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