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'm not sure hot to solve for Tf? The formula itself makes sense, I'm having trouble using it. I tried to divide one side away from the other and that didn't come up with the answer (which is 210 C).

Explanation / Answer

Let the final temperature of rivet and mercury be Tf

Change in temperature of rivet = 500 - Tf

Change in temperature of mercury = Tf - 20

Heat lost = Heat gained

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

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

=> 22400 - 44.8Tf = 69Tf - 1380

-> (69 + 44.8)Tf = 22400 + 1380

=> Tf + ( 22400 + 1380) / ( 69 + 44.8)

Tf = 208.96 degree C 210 degree C

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