# Describe what would happen to the temperature of water if a person poured 2 liters of water at 25 degrees C into a container that has 2 liters of water at 75 degrees C. What temperature would the mixture become?

Jul 8, 2017

Conceptually, we should be able to figure this out in less than $2$ minutes (i.e. this is a straightforward ACS-style question).

The final temperature for two equal masses of water combined will be halfway between the high and low temperatures.

Mathematically, we assume conservation of thermal energy out from the hot body of water into the cold body of water:

${q}_{c o l d} = m {C}_{P} \Delta {T}_{c o l d}$

$= - {q}_{h o t} = - m {C}_{P} \Delta {T}_{h o t}$,

where $q$ is heat flow, ${q}_{h o t} < 0$, and ${q}_{c o l d} > 0$. ${C}_{P} = \text{4.184 J/g"^@ "C}$ is the specific heat capacity of water at constant pressure (lab bench conditions), and $m$ is the mass of the water in $\text{g}$.

Both bodies of water have the same volume and hence approximately the same mass (more or less...), assuming similar densities at these different temperatures.

If we assume we don't know what the final temperature is, except for that it will be the same for both bodies of water (as required to reach thermal equilibrium), then:

$\cancel{m {C}_{P}} \Delta {T}_{c o l d} = - \cancel{m {C}_{P}} \Delta {T}_{h o t}$

$\implies {T}_{f} - {T}_{i , c o l d} = - \left({T}_{f} - {T}_{i , h o t}\right)$

=> T_f - 25^@ "C" = -(T_f - 75^@ "C")

$\implies {25}^{\circ} \text{C" - T_f = T_f - 75^@ "C}$

$\implies {100}^{\circ} \text{C} = 2 {T}_{f}$

$\implies$ $\textcolor{b l u e}{{T}_{f} \approx {50}^{\circ} \text{C}}$

Indeed, ${50}^{\circ} \text{C}$ is right in the middle of ${25}^{\circ} \text{C}$ and ${75}^{\circ} \text{C}$, as predicted conceptually.