Question

The temperature of a 12.58g sample of calcium carbonate increases from 23.6degrees celcius to 38.2 degrees celcius. If the heat capacity is 0.82J/g-K, how many Joules of heat are absorbed
Looks to me like the temp change is 14.6K, so

.82J/gK * 12.58g * 14.6K = 150.6J
q=m*s*t
q=(12,58)*0.82*(38,2-23.6)
q=151 Joules

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