Answer:
[tex]c_{p, sample} = 0.290\,\frac{kJ}{kg\cdot ^{\textdegree}C}[/tex]
Explanation:
The heat rejected by the sample is equal to the heat received by the water. Then:
[tex]-Q_{sample} = Q_{water}[/tex]
[tex]-(0.11\,kg)\cdot c_{p.sample} \cdot (30.56\,^{\textdegree}C-82\,^{\textdegree}C) = (0.11\,kg)\cdot (4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (30.56\,^{\textdegree}C-27\,^{\textdegree}C)[/tex]
The specific heat of the sample is:
[tex]c_{p, sample} = 0.290\,\frac{kJ}{kg\cdot ^{\textdegree}C}[/tex]
