Question

Determine the specific heat of a 70 g sample of material that absorbed 96 J as it was heated from 293 K to 313 K

Answer 1

This problem is providing us with the mass (70 g), absorbed heat (96 J) and initial and final temperatures (293 K and 313 K, respectively) so the specific heat of the material is required and found to be 0.0686 J/(g*K) as shown below:

Calorimetry:

In chemistry, we can go over calorimetry by writing the following relationship among heat, mass, specific heat and temperature change:

$Q=mC(T_f-T_i)$

Thus, one can get the specific heat by solving for C in the previous equation:

$C=\frac{Q}{m(T_f-T_i)}$

Hence, we can plug in the given data to obtain:

$C=\frac{96J}{70g(313K-293K)}\\ \\C=0.0686\frac{J}{g*K}$

Learn more about calorimetry: brainly.com/question/1407669