Answer
The equilibrium temperature will be $18.5~^{\circ}C$
Work Step by Step
Let $T_e$ be the equilibrium temperature. The heat energy lost by the copper will be equal in magnitude to the heat energy gained by the aluminum and the water. Therefore:
$m_c~c_c~(T_c-T_e) = m_w~c_w~(T_e-T_w)+m_a~c_a~(T_e-T_a)$
$T_e~(m_c~c_c+m_w~c_w+m_a~c_a) = m_c~c_c~T_c+m_w~c_w~T_w+m_a~c_a~T_a$
$T_e = \frac{m_c~c_c~T_c+m_w~c_w~T_w+m_a~c_a~T_a}{m_c~c_c+m_w~c_w+m_a~c_a}$
$T_e = \frac{(0.265~kg)(390~J/kg~C^{\circ})(245^{\circ}C)+(0.825~kg)(4186~J/kg~C^{\circ})(12.0^{\circ}C)+(0.145~kg)(900~J/kg~C^{\circ})(12.0^{\circ}C)}{(0.265~kg)(390~J/kg~C^{\circ})+(0.825~kg)(4186~J/kg~C^{\circ})+(0.145~kg)(900~J/kg~C^{\circ})}$
$T_e = 18.5~^{\circ}C$
The equilibrium temperature will be $18.5~^{\circ}C$.
