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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.

Respuesta :

Willchemistry
Q =  M * C *ΔT

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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