Posted by John on Sunday, March 21, 2010 at 10:04pm.
The amount of heat transfered to the water is 150*7.1*1.0 = 1065 calories.
Removing that amount of heat from 100 g of the substance caused its temperature to drop 67.9 C. The specific heat can be calculated from that information.
C = Q/(M*deltaT) = 0.157 Cal/gm C
= 0.657 J/g K
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