Question #f3dcd

1 Answer
Apr 8, 2017

Answer:

#"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