# If you combine 300.0 mL of water at 25.00°C and 110.0 mL of water at 95.00°C, what is the final temperature of the mixture?

Dec 23, 2016

I got ${44}^{\circ} C$ approx.

#### Explanation:

We can say that the heat lost by the hot water ${Q}_{1}$ is equal to the heat ${Q}_{2}$ gained by the cold where the heat can be written as funcion of mass $m$, specific heat $c$ and the variation of temperature between final and initial temperature $\Delta T = \left({T}_{f} - {T}_{i}\right)$.
We will have:
${Q}_{1} + {Q}_{2} = 0$
${m}_{1} \cancel{c} \left({T}_{f} - {T}_{1}\right) + {m}_{2} \cancel{c} \left({T}_{f} - {T}_{2}\right) = 0$

With $1 m l = 1 m g$ for mass:

$110 \left({T}_{f} - 95\right) + 300 \left({T}_{f} - 25\right) = 0$
$110 {T}_{f} - 10 , 450 + 300 {T}_{f} - 7 , 500 = 0$
$410 {T}_{f} = 17 , 950$
So:
${T}_{f} = \frac{17 , 950}{410} = 43.78 \approx {44}^{\circ} C$