An aluminium vessel of mass 0.5 kg holds 0.2 kg of water at 20^ ^ ^...

Using the principle of calorimetry, heat lost by the hot body is equal to the heat gained by the cold bodies.

Let the equilibrium temperature of the mixture be $T$.

Heat lost by the iron block:
$Q_{Fe} = m_{Fe} c_{Fe} (100 - T)$
$Q_{Fe} = 0.2 \times 470 \times (100 - T) = 94 (100 - T)$

Heat gained by the aluminium vessel and water:
$Q_{Al} + Q_{w} = m_{Al} c_{Al} (T - 20) + m_{w} c_{w} (T - 20)$
$Q_{Al} + Q_{w} = (0.5 \times 910 + 0.2 \times 4200) (T - 20)$
$Q_{Al} + Q_{w} = (455 + 840) (T - 20) = 1295 (T - 20)$

Equating heat lost to heat gained:
$94 (100 - T) = 1295 (T - 20)$
$9400 - 94 T = 1295 T - 25900$
$1389 T = 35300$
T = \dfrac{35300}{1389} \approx 25.4^{}^{{}^{\circ}}\text{C}