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KuraJones
KuraJones KuraJones
  • 25-02-2016
  • Chemistry
Answered

  1. 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?

Verified Answer :

Willchemistry
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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