2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, cal
culate the specific heat of aluminum.
1 answer:
Q = M * C *ΔT
Q / <span>ΔT = M
</span>Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC
<span>
C = 1137 J / 140 * 36.2
C = 1137 / 5068
C = 0.224 J/gºC</span>
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