KuraJones KuraJones

25-02-2016
Chemistry

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4. A total of 226 J of heat are absorbed as 58.3 g of lead is heated from 12.0oC to 42.0oC. From this data, what is the specific heat of lead?

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Willchemistry Willchemistry

25-02-2016

C = Q / M * ΔT

Δf - Δi = 42.0ºC - 12.0ºC = 30.0ºC

C = 226 J / 58.3 * 30.0

C = 226 / 1749

C = 0.129 J/gºC

hope this helps!

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