# What is the final temperature when 50 mL of water at 80°C are added to 25 mL of water at 25°C?

Jun 1, 2017

${61.7}^{o} C$

#### Explanation:

$\sum$ energy changes = 0

$\sum Q = {Q}_{{80}^{o} C} + {Q}_{{25}^{o} C}$ = [mcDeltaT_(80^oC) + mcDeltaT_(25^oC)] = 0

Density HOH = 1.00 g/ml;
Speific Heat(c) HOH = $1.00 \frac{\text{cal}}{g} ^ o C$

$\left[\left(50 g\right) \left(1 \left(\text{cal"/g^oC))(T_("final}\right) - {80}^{o} C\right)\right]$ +

$\left[\left(25 g\right) \left(1 \left(\text{cal"/g^oC))(T_("final}\right) - {25}^{o} C\right)\right]$ = $0$

=> $\left[\left(50\right) \left(1\right) \left({T}_{\text{final") - 80)] + [(25)(1)(T_("final}} - 25\right)\right] = 0$

=> $\left(50 {T}_{\text{final") - 4000) + (25T_("final}} - 625\right) = 0$

=> $75 {T}_{\text{final}} = \left(4000 + 625\right) = 4625$

=> $T = {\left(\frac{4625}{75}\right)}^{o} C = {61.7}^{o} C$