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, calculate the specific heat of aluminum.
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Ответ:
Q / ΔT = M
Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC
C = 1137 J / 140 * 36.2
C = 1137 / 5068
C = 0.224 J/gºC
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