Answer
The latent heat of fusion of mercury is $11.1~kJ/kg$
Work Step by Step
We can find the heat energy lost by the water and the aluminum.
$Q = m_w~c_w~\Delta T + m_a~c_a~\Delta T$
$Q = (0.400~kg)(4186~J/kg~C^{\circ})(12.80~^{\circ}C-5.06~^{\circ}C) + (0.620~kg)(900~J/kg~C^{\circ})(12.80~^{\circ}C-5.06~^{\circ}C) $
$Q = 17,278.8~J$
This is the heat energy required to melt the mercury and raise the temperature of the mercury. We can find the latent heat of fusion of mercury.
$Q = m~L + m~c_m~\Delta T$
$L = \frac{Q-m~c_m~\Delta T}{m}$
$L = \frac{17,278.8~J-(1.00~kg)(140~J/kg~C^{\circ})(5.06^{\circ}C-(-39.0^{\circ}C))}{1.00~kg}$
$L = 11,100~J/kg$
$L = 11.1~kJ/kg$
The latent heat of fusion of mercury is $11.1~kJ/kg$.
