If you combine 250.0 mL of water at 25°C and 120.0 mL of water at 95°C, what is the final temperature of the mixture?

Jul 4, 2016

${T}_{f}^{o} C = {47.07}^{o} C$

Explanation:

We can use a heat equilibrium equation to find this value.

$\Delta H = m \left({T}_{f} - {T}_{i}\right) {C}_{p}$

$\Delta H =$ Change in Heat
$m =$ Mass
$T =$ Tempurature
${C}_{p} =$ Specific Heat

$\Delta H$ cold ${H}_{2} O$ = -$\Delta H$ hot ${H}_{2} O$

$m \left({T}_{f} - {T}_{i}\right) {C}_{p} = - \left[m \left({T}_{f} - {T}_{i}\right) {C}_{p}\right]$

Since the density of water is $1 \frac{g}{m L}$ the volume of water is also equal to the mass of the water in grams.

Cold ${H}_{2} O$

$m = 250.0 g$
T_f= ?
${T}_{i} = {25}^{o} C$
${C}_{p} = 4.18 \frac{J}{{g}^{o} C}$

Cold ${H}_{2} O$

$m = 120.0 g$
T_f= ?
${T}_{i} = {95}^{o} C$
${C}_{p} = 4.18 \frac{J}{{g}^{o} C}$

$m \left({T}_{f} - {T}_{i}\right) {C}_{p} = - \left[m \left({T}_{f} - {T}_{i}\right) {C}_{p}\right]$

$250.0 g \left({T}_{f} - {25}^{o} C\right) 4.18 \frac{J}{{g}^{o} C} = - \left[120.0 g \left({T}_{f} - {95}^{o} C\right) 4.18 \frac{J}{{g}^{o} C}\right]$

$1 , 045 {T}_{^} o C - 26 , 125 \frac{J}{{g}^{o} C} = - 501.6 {T}_{f}^{o} C + 47 , 652 \frac{J}{{g}^{o} C}$

$\frac{\cancel{1 , 546.6} {T}_{f}^{o} C}{\cancel{1 , 546.6}} = \frac{73 , 777}{1 , 546.6}$

${T}_{f}^{o} C = {47.07}^{o} C$