Question

A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H O(l) at 27.00°C in an insulated container. The temperature rises to 30.56°C. Neglecting the heat capacity of the container, what is the specific heat of the metal? The specific heat of H O(l) is 4.18 J/(g ∙ °C).

Answer 1

Answer:

$c_{p, sample} = 0.290\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

Explanation:

The heat rejected by the sample is equal to the heat received by the water. Then:

$-Q_{sample} = Q_{water}$

$-(0.11\,kg)\cdot c_{p.sample} \cdot (30.56\,^{\textdegree}C-82\,^{\textdegree}C) = (0.11\,kg)\cdot (4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (30.56\,^{\textdegree}C-27\,^{\textdegree}C)$

The specific heat of the sample is:

$c_{p, sample} = 0.290\,\frac{kJ}{kg\cdot ^{\textdegree}C}$