# What would be the final temperature when 100 g of 25° C water is mixed with 75 g of 40°C water?

##### 1 Answer
Aug 14, 2016

I got about ${31.4}^{\circ} \text{C}$.

This is asking you to look at the heat transfer between a hotter substance and a colder substance, and the hotter transfers to the colder. I would expect the result to be closer to ${25}^{\circ} \text{C}$ than to ${40}^{\circ} \text{C}$ because there is more colder water than hotter water.

So, you will have to use the heat flow equation:

$\boldsymbol{q = \pm m c \Delta T = \pm m c | {T}_{f} - {T}_{i} |}$

Keep in mind that $q > 0$ if heat is absorbed and $q < 0$ if heat is released. So $q < 0$ for the hotter water and $q > 0$ for the colder water. That is so that:

${q}_{\text{hotter" + q_"colder}} = 0$, i.e. we have conservation of energy for ideal closed systems.

${q}_{\text{hotter}} = - m c \Delta T$

= -("75 g")("4.184 J/g"""^@"C")|T_f - 40^@ "C"|

$= - \left(\text{75 g")("4.184 J/g"""^@"C")(40^@ "C} - {T}_{f}\right)$
(because for the hotter water, ${T}_{f} < {40}^{\circ} \text{C}$)

${q}_{\text{colder}} = m c \Delta T$

= ("100 g")("4.184 J/g"""^@"C")|T_f - 25^@ "C"|

In an ideal closed system, like the one we assume we have, ${q}_{\text{hotter" = -q_"colder}}$.

So:

$- \left(\text{75 g")("4.184 J/g"""^@"C")(40^@ "C" - T_f) + ("100 g")("4.184 J/g"""^@"C")(T_f - 25^@ "C}\right) = 0$

$\left(\text{75 g")cancel(("4.184 J/g"""^@"C"))(40^@ "C" - T_f) = ("100 g")cancel(("4.184 J/g"""^@"C"))(T_f - 25^@ "C}\right)$

0.75(40^@ "C" - T_f) = T_f - 25^@ "C"

${30}^{\circ} \text{C" - 0.75T_f = T_f - 25^@ "C}$

$1.75 {T}_{f} = {55}^{\circ} \text{C}$

$\textcolor{b l u e}{{T}_{f} \approx {31.4}^{\circ} \text{C}}$

This makes sense because thermal equilibrium should be established in a way that balances the heat transfer between the water, and that is when both have reached a temperature in between each other's.