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A 30.73 g 30.73 g sample of a substance is initially at 20.8 ° C. 20.8 °C. After absorbing 2813 J 2813 J of heat, the temperature of the substance is 163.2 ° C. 163.2 °C. What is the specific heat ( c c ) of the substance?

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Answer:

[tex]Cp=0.643\frac{J}{g^0C}[/tex]

Explanation:

Hello,

In this case, the following hear+t equation is taken in account in terms of the required specific heat:

[tex]Q=mCp\Delta (T_2-T_1)[/tex]

Thus, as the the heat is absorbed it is taken positive, therefore, the heat capacity results:

[tex]Cp=\frac{Q}{m\Delta (T_2-T_1) }=\frac{2813J}{30.73 g(163.2-20.8)^0C} \\Cp=0.643\frac{J}{g^0C}[/tex]

Best regards.

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