The impatient hot tub owner An impatient man owns a 300 gallon hot tub. He comes home from a hard day of work and sees that his hot tub is currently simmering at 90F. For maximum relaxation, he wants it at 104F. However, the hot tub pump/heater is pretty slow: maybe 1 or 2 degrees an hour.
But this man wants the hot tub hot within the hour. Is there any way for him to heat up the water to 104F that fast (while still being safe for him to soak in)?
One idea that I had: what if he boiled a pot of water and poured it in to the tub. How much boiled water would need to do that? What would the equation for that look like? How would it affect the pH/bromine levels?
Are there other practical methods that might accomplish the same thing?
 A: Assuming no heat is lost to the environment the heat balance on adding some boiling water ($212\:\mathrm{F}$) is given by:
$$m_{bath}cT_{bath}+m_{added}cT_b=(m_{bath}+m_{added})cT_f$$
where: $m_{bath}=300\:\mathrm{Gall}$ is the initial amount of water, $T_{bath}=32.2\:\mathrm{Celsius}$, $m_{added}$ the amount of boiling water added, $T_b=100\:\mathrm{Celsius}$, $c$ the heat capacity of the water which we assume to be constant and $T_f=40\:\mathrm{Celsius}$. So we have:
$$300 \times 32.2+m_{added} \times 100 = (300+m_{added}) \times 40$$
So:
$m_{added}=39\:\mathrm{Gall}$ of water at $212\:\mathrm{F}$ added to the bath raises its temperature approx. to $104\:\mathrm{F}$.
Assuming tap water was used for the initial bath and the added water there will be no effect on $pH$ or mineral levels.

Using 'PaulT' suggestion (comments) the heat balance by replacing some of the bath water with boiling water is:

$$(300-m_{add})cT_{bath}+m_{added}cT_b=300cT_f$$
Thus: $m_{add}=34.5\:\mathrm{Gall}$.
A: It's not the question asked, but looking at the power requirements might give some insight.  Raising the water temperature requires a specific amount of energy, and the time constraint gives a required power.
$$P = \frac{m C T} { t}$$
$$P = \frac{300\text{gallon }(1000\text{kg/m^3})(4.186\text{J/g K}) 14\text{degF}}{1 \text{hour}}$$
$$P = \frac{1135\text{kg} (4186\text{J/kg K})7.8\text{K}}{3600\text{s}}$$
$$P = 10.3\text{kW}$$
That's a lot of power.  On a $230V$ circuit, it would have to pull at least $45A$.  You're not going to get that on a regular household electric stove.
For a gas stovetop, a "power burner" might be in the range of $25,000 \text{BTU}$ or about $7.3kW$, still not enough by itself.  You'd need a dedicated high-power heating unit.
A: A typical hot tub will be coming to an equilibrium based on some forcing term $F$ adding heat to the system plus some proportional response $\lambda$ which loses heat to the environment:$$\rho \frac{dT}{dt} = F - \lambda (T - T_0)$$This is a linear ODE whose equilibrium temperature is $T = T_0 + F/\lambda.$ To increase $T$ as fast as possible you should:


*

*Turn up the thermostat in general ($\uparrow T_0.$)

*Turn up the temperature regulator on the hot tub ($\uparrow F.$)

*Thermally insulate the hot tub as much as possible. ($\downarrow \lambda.$)


As for whether boiling water will work, each gallon of water boiled will contribute approximately $$(212^\circ\text{F} - 90^\circ\text{F})\cdot 1\text{ gal}/300\text{ gal} = 0.4^\circ\text F$$ to the temperature of the hot tub, getting up to about 10% less efficient as the hot tub warms up to the desired temperature. With two burners and two 16-quart stock pots making maybe 6 gallons every 15 minutes, you have about 24 gallons for about $10^\circ\text F,$ which gets you almost all the way there. The hot tub will have about 10% lower concentration of whatever salts are in it.
A: Instead of just adding boiling water as @Gert did, lets drain cool water and replace it with boiling.  This way we'll keep the total volume of water constant $V=300$ gal.
We can determine what volume of water to add $v_\mathrm{add}$.
$$(V-v_\mathrm{add})\cdot T_\mathrm{orig} + v_\mathrm{add}\cdot T_\mathrm{add} = V\cdot T_\mathrm{target}$$
$$v_\mathrm{add} = V \left(\frac{T_\mathrm{target}-T_\mathrm{orig}}{T_\mathrm{add}-T_\mathrm{orig}}\right) \approx 34.5\,\mathrm{gal}$$
We need 34.5 gal or 130 L of boiling water. Lets first drain 130 L of 90 F (305 K) water from the hot tub. Then, we'll heat that.  We need to raise the temperature 68 K to boil it.  How much energy do we need?
$$ E= m\,c\,\Delta T$$
$$ E = (130\,\mathrm{L}) \cdot \frac{1\,\mathrm{kg}}{\mathrm{L}} \cdot \frac{4180\,\mathrm{J}}{\mathrm{kg}\cdot\mathrm{K}} \cdot (68\,\mathrm{K}) \approx 37.1\,\mathrm{MJ}$$
That's 37 million Joules (about 35 000 BTU).
If we wish to soak in the hot tub in one hour, we'll need a 10.3 kW heater!
$$P=\frac{\mathrm{d}E}{\mathrm{d}t} = \frac{37.1\,\mathrm{MJ}}{3600\,\mathrm{sec}} \approx 10.3\,\mathrm{kW}$$
As @BowlOfRed points out, you won't get that kind of power from any basic household appliance.
