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I would like to calculate how long it takes to boil a pan of water on an electric stove.

The water has a mass of 1 kg, an initial temperature of 15°C and has a convective heat transfer coefficient of 4000 $W/m^2 K$. The pan has a thermal conductivity of 240 W/m K, an area of 0.0176 $m^2$ and is 0.0035 m thick. The stove supplies 2000 Watts of power and surface of the stove is at a 200°C (I have tried to make educated guesses for these values but a lot of them are either made up or are as close to real values as I could find, some of them may not make sense).

I know that the energy required to heat the water from 15°C to 100°C is given by:

$$Q=m\,Cp∆T=1\,\times4200\,\times\,(100-15)=357kJ $$

If I neglect the resistance due to convection and radiation between the stove and the pan, I think the resistance profile will look like this:

$$T_{stove}\,-\,R_{cond}\,-\,T_{surf}\,-\,R_{conv}\,-\,T_{water}$$

The total resistance is given by:

$$R_T = R_{cond} + R_{conv}=\frac{L}{kA}+\frac{1}{hA}= \frac{0.0035}{240 \times 0.0176}+\frac{1}{4000 \times 0.0176}= 0.0150\, K W^{-1}$$

Where:

$R_{cond}$ = resistance due to conduction in the base of the pan
$R_{conv}$ = resistance due to convection between the internal surface of the pan and the water

I know that the relationship between heat transfer rate and total resistance is given by:

$$ \dot{Q}=\frac{∆T}{R_T} = \frac{T_{stove}-T_{water}}{R_T}$$

I think my problem is that I don't understand the relationship between the energy required to raise the temperature of the water to 100°C, the power supplied by the stove and the total resistance. How does the resistance affect the amount of energy that the water receives? How can I find the time taken to heat the water?

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The amount of time to boil cannot be calculated simply as you intend because the water temperature changes with time.

The heat flux $\dot{Q}$ flowing from a hot source (at constant) temperature $T_H$ to a colder source $T(t)$ (not a constant) is given by Newton's Law of Cooling/Heating:

$$\dot{Q}=uA[T_H-T(t)]\tag{1}$$

where $u$ is the overall heat transfer coefficient.

In your case $u$ is made up of three elements:

  1. electric element to pan: $h_1$
  2. conduction through pan: $h_2$
  3. convection of pan to water: $h_3$

$u$ is calculated as:

$$\frac{1}{u}=\frac{1}{h_1}+\frac{L}{k}+\frac{1}{h_3}$$

Your question shows that you know the first two terms but not the last one.

Even if you did, you wouldn't be able to calculate the time to boil as you wish to do.

Look at $(1)$ and we have:

$$\dot{Q}=\frac{\text{d}Q}{\text{d}t}=-mc_p\frac{\text{d}T(t)}{\text{d}t}$$

So that with $(1)$ we have:

$$-mc_p\frac{\text{d}T(t)}{\text{d}t}=uA[T_H-T(t)]\tag{2}$$

$(2)$ is a simple ordinary differential equation (ODE) you'll have to solve to calculate your heating time.

Often we set:

$$\frac{1}{\tau}=\frac{uA}{mc_p}$$

where $\tau$ is the so-called characteristic time, so:

$$\frac{\text{d}T(t)}{\text{d}t}=-\frac{1}{\tau}[T_H-T(t)]\tag{3}$$


I see no easy way to simplify this, as the OP requests. I'll therefore provide the solution to $(3)$:

$$\ln\Big(\frac{T_H-T(t)}{T_H-T_0}\Big)=-\frac{\Delta t}{\tau}$$ where $T_0$ is the starting temperature of the water and $\Delta t$ the time to reach $T(t)$.


Ah. I see a way.

The trick here is to calculate an approximate heat flux $\dot{Q}$.

Take the heat flux at $t=0$:

$$\dot{Q}_0=uA[T_H-T_0]$$

and at $t=\Delta t$:

$$\dot{Q}_t=uA[T_H-T(t)]$$

Now take the average of the two:

$$\dot{Q}_{av}=\frac{uA[2T_H-T_0-T(t)]}{2}$$

Now we also know that:

$$mc_p[T(t)-T_0]\approx\frac{uA[2T_H-T_0-T(t)]}{2}\Delta t$$

From which $\Delta t$ can be computed.

The approximation could be rendered more accurate by including a mid-point for the heat flux: $$\dot{Q}_m=uA\Big(T_H-\frac{T_0+T_H}{2}\Big)$$

Then take the average over the $3$ heat fluxes.

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  • $\begingroup$ Ok thank you for the thorough explanation, I really appreciate it. Am I able to make simplifications to the model in order to find an answer? It is quite frustrating because this problem is a part of an assignment, and my teachers know that they haven't taught how to solve transient problems yet, and they expect us to make simplifications in order to solve it. $\endgroup$ – Acrux Dec 17 '20 at 12:09
  • $\begingroup$ I'll add a simplified version in the answer. Thanks! $\endgroup$ – Gert Dec 17 '20 at 12:27
  • $\begingroup$ Wow, I've been thinking about this for a while and I think it all just clicked for me. THANK YOU SO MUCH! $\endgroup$ – Acrux Dec 17 '20 at 14:35
  • $\begingroup$ You're welcome! $\endgroup$ – Gert Dec 17 '20 at 15:12

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