If I mix 1 unit of water at at 30C° with 1 unit of water at 60C°, is the resulting water at 45C°? I'm curious how temperatures work when mixing water. I'm not very good at physics but I'm always learning.
Let's say I've 1 gallon of water at 30C° and 1 gallon of water at 60C°, and I mix them together. What is the final temperature of mixed water? Is it the average? Intuitively I'd say it's average but I don't know physics laws exactly.
 A: When two quantities of water ($m_1$ and $m_2$) at different temperatures (resp. $T_1$ and $T_2$) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature $T$ of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the mixture):
$$m_1c_wT_1+m_2c_wT_2=(m_1+m_2)c_wT,$$
where $c_w$ is the Specific Heat Capacity of water.
Since as here $m_1=m_2$ and because $c_w$ drops out of the equation, we simply get:
$$T=\frac{T_1+T_2}{2}.$$
In your case of $30^\circ$ and $60^\circ$, $T=45^\circ$.
Edit:
Following the comments, let's use the data presented on this page to evaluate some corrections suggested.
Firstly, $1\:\mathrm{Gal}=4.546\:\mathrm{l}$ and with the densities listed for the relevant temperatures we have:
$m_1=4.526\:\mathrm{kg}$ and $m_2=4.469\:\mathrm{kg}$ and $m_1+m_2=8.995\:\mathrm{kg}$
The web page also provides values of $c_w$ at $30^\circ$ of $4.178\:\mathrm{kJ/kg^\circ C}$ and at $60^\circ$ of $4.185\:\mathrm{kJ/kg^\circ C}$.
Of course we strictly speaking we don't know $T$ and thus not $c_w$ at that temperature either. I'll assume that the influence of $T$ on $c_w$ between $30^\circ$ and $60^\circ$ is linear. Then we can determine that:
$$c_w(T)=4.171+0.000233T.$$
These data can now be inserted in the original equation:
$4.526 \times 4.178 \times 30 + 4.469 \times 4.185 \times 60 = 8.995 \times (4.171+0.000233T) \times T$
or:
$0.0021T^2+37.5T-1689.5=0$
Solving this quadratic equation the usual way we get:
$$\large{T=44.94^\circ}$$
So, barely $0.06^\circ$ difference with regards to the approximate solution. Well worth the approximation, in my opinion. While the comments are correct, a significant discrepancy between the simple and more refined model would only arise at much larger temperature differences, not a mere $30^\circ$ range.
