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A155-g sample of a unknown substance was heated from 25.0°C to 40.0°C.In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance?

User Nashon
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The answer is: (5696 J) / (155 g) / (40.0 - 25.0)°C = 2.45 J/g·°C
User Jon Marnock
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