Answer:
[tex]m_{Al} = 18.019\,g[/tex]
Explanation:
There is a heating process between an aluminium piece and water, which is described by the First Law of Thermodynamics:
[tex]Q_{Al} = - Q_{w}[/tex]
[tex]m_{Al}\cdot c_{Al}\cdot (T_{o,Al}-T) = m_{w}\cdot c_{w}\cdot (T-T_{o,w})[/tex]
The mass of the aluminium piece is therefore cleared and known variables are substituted:
[tex]m_{Al} = \frac{m_{w}\cdot c_{w}\cdot (T-T_{o,w})}{c_{Al}\cdot (T_{o,Al}-T)}[/tex]
[tex]m_{Al} = \frac{(100\,g)\cdot \left(4.184\,\frac{J}{g\cdot^{\circ}C} \right)\cdot (15^{\circ}C-10^{\circ}C)}{\left(0.900\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (144^{\circ}C-15^{\circ}C)}[/tex]
[tex]m_{Al} = 18.019\,g[/tex]
