A Hard theoretic Physics And Math problem.?

Yes the pot will reach the desired temperature as a 'high heat' stove is between 232-343 degrees C and eventually, by heat transfer, the oil will be the same temperature as the stove

Now the hard part, how long will it take?

Ti (Initial Temperature) = 25 degrees C

Tf (Final Temperature) = 180 degrees C

dV/dt = 1 L/min

V = 1L

Tstove (Temperature of Stove) = 343 degrees C

c (Specific Heat Capacity of Oil) = 1970 J/KgC

density (olive oil) = 0.918 g/cm^3 (Use to calculate mass)

Lv (Latent Heat of Vaporization of Olive Oil) = ? (Assume it doesn't evaporate)

Q = mc(T2-T1)Q = m(1970)(180-25)

I think Q can be found as the heat energy stored in the stove, but I'm not sure how to do that

Once you do find mass, you can solve for time since we have V and dV/dt

Edit: I think you can use the power (watts) of the stove to create an equation involving time, then create another equation using m, V, dV/dt in terms of time then solve the system of equations

Edit2: I think Andrew is right

Just to complete Andrew's answer:

You need 4650 W to be added to the liquid constantly (and a little bit more, to catchup the 1 liter head start you gave the pot).

His answer is that you can't do that with a 2000W heater. True enough. But nobody said you can't have a more powerful heater. The question is: would it help.

Your heater goal is to maintain itself to a given temperature. You said 300 Celsius. So it all depends on how much energy you need to stay at this temperature. That is, how much energy is lost by the heater (and gained by the oil, under the assumption that there is not loss of energy)

And that depends on the transfer of energy between heater and oil.

That transfer is U, in W/m²/K

Such as

1/U = 1/λ + 1/h₀ + 1/h₁

h₀ and h₁ being internal and external transfer interface (film coefficient between heater and pot bottom, and between bot bottom and oil).

and λ being the heat transfer coefficient of the pot bottom.

We would have some hard time to find values for h₀ and h₁, so let ignore them. But understand that any value of smaller that ∞ means that 1/U is bigger than 1/λ, so U smaller than λ.

So let's assume your pot is made of silver (that the best heat transfer you can easily get) and that h₀ and h₁ are ∞.

So U=λ=418 W/m²/K

Surface of the pot bottom is π(0.3/2)² = 0.071 m²

So you have a transfer of energy which is 30 W/K

And you need that transfer to happen until the pot reach 180 Celsius, when the heater, as you said, is at 300 Celcius.

At the end of the heating (when oil is "almost" at 180 Celsius) the energy transfer is therefore of "only" 3545 W.

And that is under very favorable assumptions

a) No energy loss at all

b) Silver pot

c) No film transfer (I mean, infinite film transfer: any energy produced is instantly transferred from heater to pot bottom, and instantly transferred from pot bottom to oil. The only obstacle is to cross the pot bottom)

d) Instant transfer inside the oil itself.

d is important. Because it means that even if you somehow increase energy transfer

(you can: you can find a hotter heater, so more temperature gradient; you can find a better matter than silver — I am sure in a world of infinite pot, you have access do kryponite, which, as everybody know, has a huge λ; or you can increase the diameter of the pot), you'd still have to solve an harder problem: the race between the heat and the oil.

In other words, timing problem: how much time it takes for the heat to go from the bottom of the pot to the surface.

That delay keeps growing, as the height of the liquid increases.

So even if input of energy was sufficient you may end up with a situation where energy take an infinite time to reach the top.

Anyway, simple consideration the last drop of oil isn't at 180 Celsius. So it will take an instant for heat to transfer from the very hot surrounding oil to that last drop. Instant during which yet another drop of oil entered the pot.

So you need a tolerance for the last drop anyway.

Let have a closer look on that

For oil, coefficient is around 0.1 W/m/K(W/m/K unit is because, the bigger the surface, the more heat you transfer, obviously. Hence the W/m²/K of interface transfer. But here, it is not just an interface: you have a distance to cover. The longer the distance, the less energy you transfer.

So W/m²/K × m = W/m/K)

Here, surface is 0.071 m². So in Wm/K, the travel of energy along the pot height is 71×10⁻⁴ Wm/K

So rule of thumb: 71×10⁻⁴/H(t) W/K to go from the bottom to the pot, where H(t) is the height of the pot

That height is such as H(t)*S=(1+1/60t)/1000, S being S=0.071m² (1000 because I need volume in m³ not liter)

So H(t) = (1+t/60)/1000S

So energy transfer from bottom to top is in the order of 0.071²×0.1×1000 / (1+t/60) = 5×10⁻¹ / (1+t/60) W/K

You see the problem: sure temperature gradient will be higher and higher (the temperature of oil at the bottom end up way over your target), but it is also harder and harder to transfer it to the top.

Assuming that, after a while, the 1+ term is negligible (if you don't problem is even harder), that difficulty to bring heat from bottom to top increases linearly. So temperature at bottom should do so as well. And it won't. Because the higher temperature at bottom get, the less energy enter the oil (because of interface transfer between 300 degrees heater and oil.

At the beginning you have only at most 1375 W that can cross the height of oil (under impossibly favorable hypothesis: gradient of temperature being as if bottom layer of oil was already 300 Celsius. While Ignoring the fact that if it was so, then, there wouldn't be any energy coming from the heater). Which is not nearly enough. And the problem is getting worse and worse.

That is really an intuitive way to see it. To be accurate differential equations are needed. But rule of thumb tells that even if you had enough energy entering the oil (and you haven't), it is an impossible race.

Lots of assumptions.  The specific heat of olive oil is given as 1.97 kj/(kg K) and the density as 0.92 giving 1.80 kj / L /K You need to heat 1/60 L/s and raise it 155 degrees.  So you need 1800*155/60 W  = 4650 W  My high heat is a max of 2000 W ie it is impossible.

Does the question state the actual heat of the stove, or just say "high heat"? If the pot can hold infinite amount of liquid, then anything is possible, and the "high heat" stove can be as hot as an exploding star, for example.