A construction worker drops a hot 100 g iron rivet at 500 degrees C into a bucke

Question

A construction worker drops a hot 100 g iron rivet at 500 degrees C into a bucket containing 500 g of mercury at 20 degrees C. Assuming that no heat is lost to the surroundings or the bucket, what is the final temperture of the rivet and mercury?
I get the workbooks answer, just not how they derived one of the components.
heat loss = heat gain
(rivet) mc delta T = mc delta T (mercury)
(0.100 kg)( 448 J/kg C)(500C-Tf) = (0.500kg)(138 J/kg C)( Tf-20C)

I don't think I'm performing the algebra correctly.
44.8 ( 500-Tf) = 69 (Tf-20)
(500-Tf) = 24.2 (Tf-20)
Now what? one Tf is neg and one pos, I know they aren't supposed to cancel out. The correct answer is 210C
Thanks!

Explanation / Answer

You are on the right track. Lets start from

(0.100 kg)( 448 J/kg C)(500C-Tf) = (0.500kg)(138 J/kg C)( Tf-20C)

Then you get...

44.8 (500 - Tf) = (69)(Tf -20)

Now distribute the 44.8 on the left and the 69 on the right to get...

22400 - 44.8Tf = 69Tf - 1380

Then add 1380 and 44.8 Tf to both sides, combine like terms and get...

23780 = 113.8Tf

Solve for Tf

Tf = 208.9 which rounds to 210 oC

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