# Question 48b2b

Jul 26, 2015

#### Answer:

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,

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