# Homework Help: - Specific Heat Problem

1. Nov 28, 2004

### porschedriver192

I have been stuck for a while, it's online (Webassign), so it told me my answer was wrong.
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A 6.3x10^2 g sample of water at 90.0°C is mixed with 4.45x10^2 g of water at 22.0°C. Assume no heat loss to the surroundings. What is the final temperature of the mixture?
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The formula I know is right:

(cp)(m)(Tf- Ti) = (cp)(m)(Tf- Ti)

cp= spec. heat
m = mass
Tf = final Temp
Ti = initial Temp
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work:

converted grams into kilograms...

cp water, given
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(4186) (.630) (x-90) = (4186) (.445) (x-22)

774.41x = 196365.26

x = <<<it says it's wrong>>>

Where did I slip up?

Thanks.

2. Nov 28, 2004

### Staff: Mentor

I think you mean:
(cp)(m1)(Tf- Ti) + (cp)(m2)(Tf- Ti) = 0

3. Nov 28, 2004

### porschedriver192

My textbook gave me the original formula...but would the one you gave yield a correct answer? Thanks.

4. Nov 28, 2004

### Staff: Mentor

Are you sure you copied it exactly? The formula you gave makes no sense. Note that the final temperature will be somewhere between 22°C and 90°C. Thus one side of your equation would be negative, the other positive. Nothing good can come of that.

5. Nov 28, 2004

### porschedriver192

Doc Al,

I called a classmate, and he had used the same formula I did...and got the correct answer. After walking through my steps with him, I found that messed up here:

cp water, given
||
(4186) (.630) (x-90) = (4186) (.445) (x-22)

774.41x = 196365.26
It should have been 90-x
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About the equation...my book is right. It is on specific heat capactity, and states how the energy absorbed by one object must be released by the other, similar to conservation of momentum.

Thanks for the help, and I was eventually led in the right direction!

6. Nov 28, 2004

### Staff: Mentor

Well... that formula is still wrong for the reasons I gave. (I'm guessing that you copied it incorrectly.)
Right! I hope you realize that since you had to reverse things in your formula, that your formula cannot be correct.