Question #bfecf

1 Answer
Nov 4, 2016

Answer:

#"2140 J"#

Explanation:

For starters, let's figure out if the heat is evolved, which is a fancy term used to mean given off, or absorbed.

Notice that the temperature of the water decreases from #89.9^@"C"# to #76^@"C"#, which can only mean that heat was given off.

Now, look at the specific heat of water, which is given to you as #"4.18 J g"^(-1)"K"^(-1)#. This tells you that in order to increase the temperature of #"1 g"# of water by #"1 K"#, which is equivalent to increasing the temperature by #1^@"C"#, you must provide it with #"4.18 J"# of heat.

Similarly, when the temperature of #"1 g"# of water decreases by #1^@"C"#, #"4.18 J"# of heat are being given off.

You can thus say that the specific heat of water is

#"4.18 J g"^(-1)"K"^(-1) = "4.18 J g"^(-1)""^@"C"^(-1) = "4.18 J"/("1 g" * 1""^@"C")#

You can use the specific heat of water to calculate how much heat would be given off when the temperature of #"1 g"# of water decreases by

#DeltaT = |76^@"C" - 89.9^@"C"| = 13.9^@"C"#

In this case, you would have

#13.9 color(red)(cancel(color(black)(""^@"C"))) * "4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "58.102 J g"^(-1)#

So, you know that when #"1 g"# of water cools by #13.9^@"C"#, #58.102 J"# of heat are being given off. Now all you have to do is figure out how much heat would be given off by your #"36.8-g"# sample

#36.8 color(red)(cancel(color(black)("g"))) * "58.102 J"/(1color(red)(cancel(color(black)("g")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("2140 J")color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.