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)

since c = Q/ m delta T I don't understand how they derived c for each side of the equation? Thanks!

Explanation / Answer

The c values are called the specific heat. They are constants and not derived. The values are most likely found in a table in your book. For example, the specific heats of some substances are as follows

Aluminum - 900 J/kg oC

Beryllium - 1820 J/kg oC

Copper - 387 J/kg oC

Glass - 837 J/kg oC

Gold - 129 J/kg oC

Iron - 448 J/kg oC

Lead - 128 J/kg oC

Mercury - 138 J/kg oC

Silver - 238 J/kg oC

Water - 4186 J/kg oC

Every element or compound has a specific heat dependent on the nature of the atomic structure. It is not a derived value, but a property of the compound. They are known. Check your book in the chapter you are in for a table of these constants.

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