Question #bfecf

1 Answer
Stefan V.
Nov 4, 2016

"2140 J"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"89.9∘C to 76^@"C"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)4.18 J g−1K−1. This tells you that in order to increase the temperature of "1 g"1 g of water by "1 K"1 K, which is equivalent to increasing the temperature by 1^@"C"1∘C, you must provide it with "4.18 J"4.18 J of heat.

Similarly, when the temperature of "1 g"1 g of water decreases by 1^@"C"1∘C, "4.18 J"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")4.18 J g−1K−1=4.18 J g−1∘C−1=4.18 J1 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"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.

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