An iron cube (density = 8000 kg m^ -3 , specific heat capacity =...

Let the volume of the iron cube be $V$. Since the upper surface of the cube just aligns with the surface of the ice at the final equilibrium position, the volume of the ice melted is exactly equal to the volume of the cube.

Mass of the ice melted is given by:
$m_{ice} = \rho_{ice} V$

The heat required to melt this amount of ice is:
$Q = m_{ice} L_f = \rho_{ice} V L_f$

The heat released by the iron cube as it cools from its initial temperature $T$ to 0^{}^{{}^{\circ}} \text{C} is:
$Q = m_{Fe} c_{Fe} T = \rho_{Fe} V c_{Fe} T$

By the principle of calorimetry, assuming no heat loss to the surroundings, the heat lost by the iron cube equals the heat gained by the ice:
$\rho_{Fe} V c_{Fe} T = \rho_{ice} V L_f$

Solving for the initial temperature $T$:
$T = \dfrac{\rho_{ice} L_f}{\rho_{Fe} c_{Fe}}$

Substituting the given values:
$T = \dfrac{900 \times 3.36 \times 10^5}{8000 \times 470}$

$T = \dfrac{9 \times 336000}{80 \times 470}$

T = \dfrac{302400}{3760} \approx 80.4^{}^{{}^{\circ}} \text{C}