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Chemistry

Answered

A 40.5 g sample of aluminum is cooled 7.5°C. The specific heat capacity of aluminum is
0.900 J/g°C. What is the energy change for this sample?

Answer :

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Q=mc\Delta T
Q= 40.5g x 0.900 J/g degree Celsius x 7.5
Q= 273.4 J.
Even if it is cooled , we use the same temperature as heating it.

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