Question

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?

Answer 1

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!