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Math used in this equation rearrangement?

  1. Dec 10, 2014 #1
    I'm trying to determine how much water it takes to raise a car's temperature from -25°C to 0°C. The water is at 10°C.

    What I apparently need to have set up is:

    -ΔUint,water = ΔUint,car

    -(mwcwΔTw) - mwLf,w = mcccΔTc

    The resultant rearranged equation looking for mass of water gives this:

    mw = -(mcccΔTc)/cwΔTw - mwLf,w

    I don't understand how this was done. Also, why are you subtracting Latent heat of fusion x Mass from mwcwΔTw?
     
  2. jcsd
  3. Dec 10, 2014 #2

    NascentOxygen

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    Looks like there's a mistake in the algebra, then. What do you think that eqn should be?

    Q for you: why does latent heat of fusion enter into the picture at all?
     
  4. Dec 10, 2014 #3

    jtbell

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    Says who?

    (Hint: are the units consistent?)
     
  5. Dec 10, 2014 #4
    Latent heat of fusion is in the picture because the water is going from a liquid to a solid. It freezes when it hits the car, then at a certain point it doesn't because the car warms up to 0 degrees. And I don't know what's up with the equation, but that's what my professor's solution says. Based on my math, I got something that makes no sense:

    0= mcarccarΔTcar/CwΔTw + Lf
     
  6. Dec 11, 2014 #5

    NascentOxygen

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    The left side should be mw. You need a pair of brackets on the right side, and then it should look right.
     
  7. Dec 11, 2014 #6
    I don't understand how. Is there any way (and I know this is no easy task) to break down the algebra step by step for me?
    Also, I made a mistake with the resultant equation in my first post. It's actually mw = -(mcccΔTc)/cwΔTw - Lf,w (no - mwLf,w)

    I realize that ends up being the same thing that you said and there is no error (but he kept the negatives in, whereas we cancelled them out), but I still don't understand how.
     
  8. Dec 12, 2014 #7

    NascentOxygen

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    You started with this: -mwcwΔTw - mwLf,w = mcccΔTc
    Taking out a common factor -mw we have
    -mw(cwΔTw +Lf,w)= mcccΔTc

    Now divide both sides by (cwΔTw +Lf,w)
    and we are left with
    -mw = mcccΔTc / (cwΔTw +Lf,w)

    Multiplying both sides by -1 so that we end up with mw by itself,
    mw = -mcccΔTc / (cwΔTw +Lf,w)

    The brackets I said you needed are those in the denominator; the ones you added in the numerator make no difference.
     
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