# What is the final temperature when 100 g of water at 25"^oC is mixed with 75 g of water at 50"^oC?

Oct 10, 2016

You must remeber than total energy remains constant: the thermic energy that wins the cold water must be equal to the thermic energy that loses the hot one.

#### Explanation:

The heat exchanging of a system with the outside when a mass $m$ of a material having a specific heat ${C}_{e}$ undergoes a temperature variation $\Delta t$ is

$Q = m \cdot {C}_{e} \cdot \Delta t$

Then, if we mix a mass ${m}_{1}$ of water (specific heat C_e (H_{2}O) = 4,18 kJ cdot kg^{- 1} cdot ºC^{- 1} ) at initial temperature ${t}_{1}$ with a second quantity of water, with mass ${m}_{2}$ and temperature ${t}_{2}$, the hot water will cool while the cold will warm until both reach the same temperature. Asume ${t}_{1} > {t}_{2}$ and suppose that the final temperature of the system is t. Then:

${m}_{1} \cdot {C}_{e} \left({H}_{2} O\right) \cdot \left({t}_{1} - t\right) = {m}_{2} \cdot {C}_{e} \left({H}_{2} O\right) \cdot \left(t - {t}_{2}\right)$

Substituting the values we have:

75 cdot (50 - t) = 100 cdot (t - 25) rArr 3750 - 75 t = 100 t - 2500 rArr t = 35,7ºC