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OpenStudy (jmoney132):
Suppose a 14-gram sample of iron is heated from 20.0°C to 25.0°C. The specific heat of iron is 0.11 cal/g°C. How much heat energy was absorbed by the iron? 38.5 cal 7.7 cal 636 cal 69.3 cal
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OpenStudy (sweetburger):
I think you use Delta Q = mcDeltaT So your heat absorbed is your Q, m is mass, c is specific heat, and Delta T is change in Temperature so just plug in the values
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