# If you combine 360.0 mL of water at 25.00°C and 120.0 mL of water at 95.00°C, what is the final temperature of the mixture?

##### 1 Answer

#### Answer:

#### Explanation:

The answer to this problem depends on whether or not you should approximate the **density** of water to be equal to

Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density *varies with temperature*, and that the value **only an approximation**.

The idea here is that the **heat lost** by the hot water sample will be **equal to** the **heat absorbed** by the room-temperature water sample.

#color(blue)(-q_"lost" = q_"absorbed")" " " "color(orange)("(*)")#

The *minus sign* is used here because **heat lost** carries a negative sign.

Your go-to equation here will be

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

Since you're dealing with two samples of water, you don't need to know the value of water's **specific heat** to solve for the final temperature of the mixture,

So, the *change in temperature* for the two samples will be

#"For the hot sample: " DeltaT_"hot" = T_f - 95.00^@"C"#

#"For the warm sample: " DeltaT_"warm" = T_f - 25.00^@"C"#

If you take the density to be equal to

#360.0 color(red)(cancel(color(black)("mL"))) * "1.0 g"/(1color(red)(cancel(color(black)("mL")))) = "360.0 g" " "# and#" "120.0color(red)(cancel(color(black)("mL"))) * "1.0 g"/(1color(red)(cancel(color(black)("mL")))) = "120.0 g"#

Use equation

#overbrace(-120.0color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(c_"water"))) * (T_f - 95.00^@"C"))^(color(purple)("heat lost by the hot sample")) = overbrace(360.0color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(c_"water"))) * (T_f - 25.00)^@"C")^(color(blue)("heat gained by warm sample"))#

This will get you

#-120.0 * T_f + 11400^@"C" = 360.0 * T_f - 9000^@"C"#

#480.0 * T_f = 20400^@"C"#

#T_f = (20400^@"C")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@"C"color(white)(a/a)|)))#

The answer is rounded to four **sig figs**.

**SIDE NOTE** *If you use the actual densities of water at* *and* *you will end up with a different answer*

#T_f = 38.76^@"C"#

*As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here*

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html