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?
1 answer:
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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