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

`c_(Al) = 0.89\,(J)/(kg\cdot ^(\circ)C)`

Step-by-step 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) = (Q)/(m\cdot \Delta T)`

`c_(Al) = (106.8\,J)/((10\,g)\cdot (12^(\circ)C))`

`c_(Al) = 0.89\,(J)/(kg\cdot ^(\circ)C)`