Question

Question #f9d73

Answer 1

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^@"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^@"C"` to solid at `660.4^@"C"`;

You know that

`q_1 = m * c * DeltaT`, where

`q_1` - the heat supplied to the metal;
`m` - the mass of the metal;
`c` - the specific heat of the metal;
`DeltaT` - the change in temperature, defined as the final temperature minus the initial temperature.

Plug in your values and solve for `q_1`.

`q_1 = 10.0cancel("g") * 0.89"J"/(cancel("g") ^@cancel("C")) * (660.4-580.0)cancel("K")`

`q_1 = "715.7 J"`

• Go from solid at `660.4^@"C"` to molten at `660.4^@"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 * DeltaH_"fus"`, where

`DeltaH_"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_"total" = q_1 + q_2`

`q_"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.