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A 22.44g sample of iron absorbs 180.8J of heat, upon which the tempature of the sample increases from 21.1 C to 39.0 C. What is the specific heat of iron?

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q = mass x specific heat x (Tfinal-Tinitial).
Solve for specific heat.

Q=mc(t2-t1)
180.8=22.44 x c(39.0-21.1)
c= 0.450 Jg-1.C-1

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