Question

Question #f3dcd

Answer 1

`"0.9000 J g"^(-1)""^@"C"^(-1)`

Explanation:

A substance's specific heat tells you the amount of heat needed to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

In your case, you know that it takes `"132.8 J"` of heat to increase the temperature of `"11.17 g"` of aluminium by

`DeltaT = 28.94^@"C" - 15.73^@"C" = 13.21^@"C"`

The first thing to do here is to figure out how much heat is needed in order to increase the temperature of `"11.17 g"` of aluminium by `1^@"C"`.

`1 color(red)(cancel(color(black)(""^@"C"))) * overbrace("132.8 J"/(13.21color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("true for 11.17 g of Al")) = "10.053 J"`

So, you know that if you deliver `"10.053 J"` of heat to `"11.17 g"` of aluminium, the temperature of the sample will increase by `1^@"C"`.

In order to increase the temperature of `"1 g"` of aluminium by `1^@"C"`, you will need

`1 color(red)(cancel(color(black)("g"))) * overbrace("10.053 J"/(11.17color(red)(cancel(color(black)("g")))))^(color(blue)("true for a 1"^@"C increase in temperature")) = "0.9000 J"`

Therefore, you can say that the specific heat of aluminium is equal to

`color(darkgreen)(ul(color(black)(c_"Al" = "0.9000 J g"^(-1)""^@"C"^(-1))))`

You need to supply `"0.9000 J"` of heat to increase the temperature of `"1 g"` of aluminium by `1^@"C"`.

The answer is rounded to four sig figs.

This is an excellent result because aluminium's specific heat is listed as being equal to what we found here.

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html