Question

Question #bfecf

Answer 1

`"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.