Question #f3dcd

1 Answer
Apr 8, 2017

"0.9000 J g"^(-1)""^@"C"^(-1)0.9000 J g1C1

Explanation:

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

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