# 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?

Mar 12, 2016

${42.50}^{\circ} \text{C}$

#### Explanation:

The answer to this problem depends on whether or not you should approximate the density of water to be equal to ${\text{1.0 g mL}}^{- 1}$.

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 ${\text{1.0 g mL}}^{- 1}$ is 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.

$\textcolor{b l u e}{- {q}_{\text{lost" = q_"absorbed")" " " "color(orange)("(*)}}}$

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

Your go-to equation here will be

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained / lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the 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, ${T}_{f}$.

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

$\text{For the hot sample: " DeltaT_"hot" = T_f - 95.00^@"C}$

$\text{For the warm sample: " DeltaT_"warm" = T_f - 25.00^@"C}$

If you take the density to be equal to ${\text{1.0 g mL}}^{- 1}$, then the two volumes are equivalent to

360.0 color(red)(cancel(color(black)("mL"))) * "1.0 g"/(1color(red)(cancel(color(black)("mL")))) = "360.0 g" " " and $\text{ "120.0color(red)(cancel(color(black)("mL"))) * "1.0 g"/(1color(red)(cancel(color(black)("mL")))) = "120.0 g}$

Use equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ top write

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

$- 120.0 \cdot {T}_{f} + {11400}^{\circ} \text{C" = 360.0 * T_f - 9000^@"C}$

$480.0 \cdot {T}_{f} = {20400}^{\circ} \text{C}$

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 ${25.00}^{\circ} \text{C}$ and ${95.00}^{\circ} \text{C}$, you will end up with a different answer

${T}_{f} = {38.76}^{\circ} \text{C}$

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