Question #6493b

1 Answer
Sep 15, 2016

Answer:

#"0.65 kcal"#

Explanation:

The first thing to do here is to make sure that you have a clear understanding of what a substance's specific heat tells you.

Specific heat is defined as the amount of heat needed to raise the temperature of #"1 g"# of a given substance by #1^@"C"#.

In your case, aluminium is said to have a specific heat of #"0.90 J g"^(-1)""^@"C"^(-1)#. This means that if you start with #"1.0 g"# of aluminium, you can raise its temperature by #1^@"C"# by adding #"0.90 J"# of heat.

Now, your sample contains more than #"1 g"# of aluminium. The first thing to do here is to figure out how much heat is needed to increase the temperature of #"78 g"# of aluminium by #1^@"C"#.

To do that, use the specific heat!

#78 color(red)(cancel(color(black)("g"))) * overbrace("0.90 J"/(1color(red)(cancel(color(black)("g"))) ""^@"C"))^(color(darkgreen)("the specific heat of aluminium")) = "70.2 J"""^@"C"^(-1)#

This means that in order to increase the temperature of #"78 g"# of aluminium by #1^@"C"#, you must provide #"70.2 J"# of heat. How about if you want to increase its temperature by

#"temperature increase" = 89^@"C" - 50^@"C" = 39^@"C"#

how much heat would be needed in this case?

#39 color(red)(cancel(color(black)(""^@"C"))) * overbrace("70.2 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(darkgreen)("for 78 g of aluminium")) = "2737.8 J"#

Finally, you must convert this to kilocalories by using

#color(purple)(bar(ul(|color(white)(a/a)color(black)("1 cal " = " 4.18 J")color(white)(a/a)|)))" "# and #" "color(purple)(bar(ul(|color(white)(a/a)color(black)("1 kcal " = 10^3"cal")color(white)(a/a)|)))#

You will have

#2737.8 color(red)(cancel(color(black)("J"))) * (1color(red)(cancel(color(black)("cal"))))/(4.18color(red)(cancel(color(black)("J")))) * "1 kcal"/(10^3color(red)(cancel(color(black)("cal")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.65 kcal")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the initial temperature of the sample, i.e. #50^@"C"#.