Question

How much heat must be added to 0.45 kg of aluminum to change it from its solid state at 130℃ to the liquid state at 660℃ (its melting point)? The latent heat of fusion of aluminum is 4.0 x 10⁵J / kg.

Answer 1

The total amount of heat which must be added to the aluminum in its solid phase to liquefy it is,

$$ \begin{aligned} Q_{\text {total }} &=Q_{1}+Q_{2} \\ &=m c \Delta T+m L_{f} \\ &=m\left(c \Delta T+L_{f}\right) \\ &=(0.45 \mathrm{~kg})\left[\left(9.0 \times 10^{2} \mathrm{~J} / \mathrm{kg} \cdot{ }^{0} \mathrm{C}\right)\left(660^{\circ} \mathrm{C}-130^{\circ} \mathrm{C}\right)+4.0 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right] \\ &=3.9465 \times 10^{5} \mathrm{~J} \end{aligned} $$