Question

The specific heat of aluminum is 0.214 cal/g°c. What is the energy, in calories, necessary to raise the temperature of a 55.5 g piece of aluminum from 23.0 to 48.6°C?

Answer 1

`"304 cal"`

Explanation:

The specific heat of aluminium tells you the amount of energy needed to increase the temperature of `"1 g"` of aluminium by `1^@"C"`.

`c_ "Al" = "0.214 cal g"^(-1)""^@"C"^(-1)`

You can thus say that in order to increase the temperature of `"1 g"` of aluminium by `1^@"C"`, you need to supply it with `"0.214 cal"` of heat.

Now, you know that your sample has a mass of `"55.5 g"`. Use the specific heat of aluminium to calculate how much heat would be needed to increase the temperature of this sample

`55.5 color(red)(cancel(color(black)("g"))) * "0.214 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "11.877 cal"""^@"C"^(-1)`

So, you now know that in order to increase the temperature of `"5.5 g"` of aluminium by `1^@"C"`, you need to supply it with `"11.877 cal"` of heat.

But since you know that the temperature change is equal to

`48.6^@"C" - 23.0^@"C" = 25.6^@"C"`

you can say that you will need

`25.6 color(red)(cancel(color(black)(""^@"C"))) * overbrace("11.877 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 55.5 g of Al")) = color(darkgreen)(ul(color(black)("304 cal")))`

The answer is rounded to three sig figs.