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

Feb 28, 2016

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:

(1xx32+4xx20)/5=112/5=22.4ºC

Feb 28, 2016

${22.4}^{\circ} \text{C}$

#### 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 ${20}^{\circ} \text{C}$ than to ${32}^{\circ} \text{C}$ because you have more* warm water than room-temperature water.

The equation to use here is

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

$q$ - the amount of heat gained / lost
$m$ - the mass of the sample
$c$ - the specific heat of water
$\Delta T$ - the 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 ${m}_{\text{warm}}$ to be the mass of the warmer sample, you will have

${m}_{\text{cold" = 4 * m_"warm}}$

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

$- {q}_{\text{warm" = q_"cold}}$

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

If you take ${T}_{f}$ to be the final temperature of the mixture, you can say that

-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

$- {T}_{f} + {32}^{\circ} \text{C" = 4 * T_f - 80^@"C}$

Rearrange to solve for ${T}_{f}$

${T}_{f} = \left({\left(32 + 80\right)}^{\circ} \text{C")/5 = color(green)(22.4^@"C}\right)$

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.