# Question f9d73

Jun 1, 2015

You'd need 4.7 kJ of heat to convert that much solid aluminium to fully molten at its melting point.

You start with a sample of solid aluminium at ${580.0}^{\circ} \text{C}$. In order to get the aluminium from solid to molten, you're going to have to supply enough heat to get it

• Go from solid at ${580.0}^{\circ} \text{C}$ to solid at ${660.4}^{\circ} \text{C}$;

You know that

${q}_{1} = m \cdot c \cdot \Delta T$, where

${q}_{1}$ - the heat supplied to the metal;
$m$ - the mass of the metal;
$c$ - the specific heat of the metal;
$\Delta T$ - the change in temperature, defined as the final temperature minus the initial temperature.

Plug in your values and solve for ${q}_{1}$.

${q}_{1} = 10.0 \cancel{\text{g") * 0.89"J"/(cancel("g") ^@cancel("C")) * (660.4-580.0)cancel("K}}$

${q}_{1} = \text{715.7 J}$

• Go from solid at ${660.4}^{\circ} \text{C}$ to molten at ${660.4}^{\circ} \text{C}$.

This time, the sample is undergoing a phase change, i.e. it goes from solid to molten. This transition happens at constant temperature, so the equation you're going to use is

${q}_{2} = m \cdot \Delta {H}_{\text{fus}}$, where

$\Delta {H}_{\text{fus}}$ - the enthalpy of fusion;

Once again, plug your values into the equation and solve for ${q}_{2}$

q_2 = 10.0cancel("g") * 398"J"/cancel("g") = "3980 J"

The total heat required will be equal to

${q}_{\text{total}} = {q}_{1} + {q}_{2}$

${q}_{\text{total" = 715.7 + 3980 = "4695.7 J}}$

Rounded to two sig figs and expressed in kJ, the answer will be

q_"total" = color(green)("+4.7 kJ")#

SIDE NOTE The + sign shows that the heat is supplied to the metal.