# What would be the final temperature of a mixture of 50g of water at 20c temperature and 50g of water at 40c temperature?

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10
Jul 26, 2015

30

#### Explanation:

[20+40]/2
because both have the same mass you take the average

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7
Jul 26, 2015

Final temperature: ${30}^{\circ} \text{C}$.

#### Explanation:

The idea of this problem is that the water with the higher temperature will lose heat and that the water with the lower temperature will gain the same amount of heat.

Mathematically, this is expressed as

${q}_{\text{hot" = -q_"cold}}$, where

${q}_{\text{hot}}$ - the heat lost by the warmer sample;
$- {q}_{\text{cold}}$ - the heat gained by the cooler sample.

Heat lost by a system is negative, while heat gained by a system is positive, hence the minus sign used in the above equation.

The equation that establishes a relationship between heat gained or lost and change in temperature looks like this

$q = m \cdot c \cdot \Delta T$, where

$m$ - the mass of the sample;
$c$ - the specific heat of the substance, in your case of water;
$\Delta T$ - the change in temperature, defined as ${T}_{\text{final" - T_"initial}}$

So, you mix the two samples together and a thermal equilibrium is reached, meaining that the final temperature will be the same for both samples.

This means that you can write

$m \cdot c \cdot \Delta {T}_{\text{hot" = -m * c * DeltaT_"cold}}$

which is equivalent to

$\cancel{m} \cdot \cancel{c} \cdot \Delta {T}_{\text{hot" = -cancel(m) * cancel(c) * DeltaT_"cold}}$

color(blue)(T_f) - 40^@"C" = -(color(blue)(T_f) - 20^@"C")

$\textcolor{b l u e}{{T}_{f}} - 40 = - \textcolor{b l u e}{{T}_{f}} + 20$

Finally,

$2 \textcolor{b l u e}{{T}_{f}} = 20 + 40 \implies \textcolor{b l u e}{{T}_{f}} = \left({60}^{\circ} \text{C")/2 = color(green)(30^@"C}\right)$

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1
Jul 26, 2015

The mixture temperature will be ${30}^{\circ} \text{C}$ that's why the substances are the same, the mass amounts are the same and there is no phase change. This value is also the arithmetic mean of the initials.

#### Explanation:

M grams of water with M grams of - again water - are mixing. We assume to be waited till the thermodynamic equilibrium is reached. So this means that we expect to measure the same temp ${T}_{\text{eq}}$ everywhere. The heat lost by the "hotter" amount of water will be gained by the "cooler" amount. So let's write: ${Q}_{g} = - {Q}_{L}$.

Any system gaining $Q$ joules of of energy will change it's temperature from initial ${T}_{i}$ to final ${T}_{f}$ by the relationship $Q = m . {c}_{v} . \left({T}_{f} - {T}_{i}\right)$ in which ${c}_{v}$ is the specific heat for the substance in constant specific volume (may be called density here).

Let's say M = 50g in this example. Also the specific heats are ${c}_{v}$ in open notation. Somethings will happen, heat will be transferred between water molecules and the mixture will reach ${T}_{\text{eq}}$. Let's write than :

${Q}_{g} = - {Q}_{L}$ /1/

$M . {c}_{v} . \left\{{T}_{\text{eq}} - \left(20 + 273 , 15\right)\right\} =$
$- M . {c}_{v} . \left\{{T}_{\text{eq}} - \left(40 + 273 , 15\right)\right\}$ /2/

$\cancel{M . {c}_{v}} . \left\{{T}_{\text{eq}} - \left(20 + 273 , 15\right)\right\} =$
$- \cancel{M . {c}_{v}} . \left\{{T}_{\text{eq}} - \left(40 + 273 , 15\right)\right\}$ /3/

canceling and eliminating;

$2 {T}_{\text{eq"=-20-cancel(273,15)+40+cancel(273,15)=60^@"C}}$ /4/

that's

${T}_{\text{eq" = 30^@"C}}$ /5/

will be the final and thermodynamic equilibrium temperature of the mixture in a steady state manner.

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