# What would be the final temperature if you mixed a liter of 32°C water with 4 liters of 20°C water?

##### 2 Answers

Admitting that the calorific capacity doesn't change too much and since there is no change of state, the final temperature will be the weighted average:

#### Explanation:

The idea here is that the amount of heat **lost** by the warmer sample will be equal to the amount of heat **gained** by the colder sample.

Even without doing any calculations, you should be able to predict that the final temperature of the mixture will be *closer to ** warm water than room-temperature water.

The equation to use here is

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

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

Assuming that water's *density* is constant for both samples, the ratio of *volumes* will be equivalent to the ratios of *masses*. If you take

#m_"cold" = 4 * m_"warm"#

Since the heat lost by the warmer sample is **equal** to the heat gained by the colder sample, you will have

#-q_"warm" = q_"cold"#

Here the *minus sign* is used because **heat lost** carries a negative sign.

If you take

#-overbrace(color(red)(cancel(color(black)(m_"warm"))) * color(red)(cancel(color(black)(c))) * (T_f - 32^@"C"))^(color(blue)("heat lost by the warmer sample")) = overbrace(4 * color(red)(cancel(color(black)(m_"warm"))) * color(red)(cancel(color(black)(c))) * (T_f - 20^@"C"))^(color(purple)("heat gained by the colder sample"))#

This will get you

#-T_f + 32^@"C" = 4 * T_f - 80^@"C"#

Rearrange to solve for

#T_f = ((32 + 80)^@"C")/5 = color(green)(22.4^@"C")#

You *should* round this off to one **sig fig**, since that's how many sig figs you have for the volumes of the two samples, but I'll leave it rounded to three sig figs.