Math used in this equation rearrangement?

  • Thread starter MarchON
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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?
 
  • #3
jtbell
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The resultant rearranged equation looking for mass of water gives this:

mw = -(mcccΔTc)/cwΔTw - mwLf,w
Says who?

(Hint: are the units consistent?)
 
  • #4
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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
 
  • #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.
 
  • #6
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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.
 
  • #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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