Question

An aluminum can weighing 10 g absorbs 106.8 J of heat and warms by 12 degrees C. What is the specific heat of the aluminum can? Use this scenario to answer questions 6 - 10.

Answer 1

Answer:

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

Explanation:

The heating process is modelled after the First Law of Thermodynamics:

$Q = m \cdot c_{Al}\cdot \Delta T$

The specific heat of the aluminium can is:

$c_{Al} = \frac{Q}{m\cdot \Delta T}$

$c_{Al} = \frac{106.8\,J}{(10\,g)\cdot (12^{\circ}C)}$

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$