# An unknown volume of water at 18.2°C is added to 33.5 mL of water at 35.0°C. If the final temperature is 23.5°C, what was the unknown volume?

Nov 17, 2015

$\text{72.4 mL}$

#### Explanation:

Before doing any calculations, try to predict what you would expect the answer to be.

Notice that the final temperature is closer to the temperature of the colder water sample than it is to the temperature of the warmer sample.

This means that you can expect the volume of the colder sample to be bigger than that of the warmer sample.

Now, you will need to use the density of water to determine the mass of the warmer water sample. You can find the density of water at different temperature here

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

So, at ${35.0}^{\circ} \text{C}$, water has a density of approximately $\text{0.994 g/mL}$. This means that the warmer sample has a mass of

33.5color(red)(cancel(color(black)("mL"))) * "0.994 g"/(1color(red)(cancel(color(black)("mL")))) = "33.3 g"

Now, your tool of choice will be the equation that establishes a relationship between heat gained or lost and change in temperature

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - the amount of heat
$m$ - the mass of the sample
$c$ - its specific heat, in your case equal to 4.18"J"/("g" ""^@"C")
$\Delta T$ - the change in temperature, defined s final temperature minus initial temperature

The key to this problem is the fact that the heat lost by the warmer sample will be equal to the heat gained by the colder sample.

$- {q}_{\text{lost" = q_"gained}}$

$- {m}_{\text{cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold" = m_"warm" * color(red)(cancel(color(black)(c))) * DeltaT_"warm}}$

${m}_{\text{cold" = -(DeltaT_"warm")/(DeltaT_"cold") * m_"warm}}$

The minus sign is needed because heat lost carries a negative sign as well.

Plug in your values into the above equation and solve for ${m}_{\text{cold}}$

${m}_{\text{cold" = -( (23.5 - 35.0)color(red)(cancel(color(black)(""^@"C"))))/((23.5 - 18.2)color(red)(cancel(color(black)(""^@"C")))) * "33.3 g}}$

${m}_{\text{cold" = 2.1698 * "33.3 g" = "72.25 g}}$

To get he volume of the colder sample, use water's density at ${18.2}^{\circ} \text{C}$, which is approximately $\text{0.99856 g/mL}$

72.25color(red)(cancel(color(black)("g"))) * "1 mL"/(0.99856color(red)(cancel(color(black)("g")))) = "72.353 mL"

Rounded to three sig figs, the answer will be

V_"cold" = color(green)("72.4 mL")

Indeed, the initial prediction turned out to be correct, the volume of the colder sample was indeed bigger than that of the warmer sample.