# 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?

##### 1 Answer

#### Answer:

#### 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

#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

#color(blue)(q = m * c * DeltaT)" "# , where

*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_"lost" = q_"gained"#

#-m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold" = m_"warm" * color(red)(cancel(color(black)(c))) * DeltaT_"warm"#

#m_"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_"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_"cold" = 2.1698 * "33.3 g" = "72.25 g"#

To get he *volume* of the colder sample, use water's density at

#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.