A 7 g metal sample absorbed 211 J of heat as it was heated from 28° C to 85° C. What is the specific heat of the metal?
Q = 211 J
m = 7g
ΔT\Delta TΔT = 85⁰C - 28⁰C = 57⁰C
c = specific heat
Q = mcΔT\Delta TΔT
211 = 7×c×57
c = 211/(57×7)
c = 0.529 j/g ⁰C
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