# Question bfecf

Nov 4, 2016

#### Answer:

$\text{2140 J}$

#### 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 ${89.9}^{\circ} \text{C}$ to ${76}^{\circ} \text{C}$, which can only mean that heat was given off.

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

Similarly, when the temperature of $\text{1 g}$ of water decreases by ${1}^{\circ} \text{C}$, $\text{4.18 J}$ 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 $\text{1 g}$ of water decreases by

$\Delta T = | {76}^{\circ} \text{C" - 89.9^@"C"| = 13.9^@"C}$

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 $\text{1 g}$ of water cools by ${13.9}^{\circ} \text{C}$, 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 $\text{36.8-g}$ 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.