Question

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?

Answer 1

Answer:

$Cp=0.643\frac{J}{g^0C}$

Explanation:

Hello,

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

$Q=mCp\Delta (T_2-T_1)$

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

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

Best regards.