How can I determine the coefficient $k$ in $ \dfrac{dT}{dt} = -k(T - 100 \mathrm{^\circ C}) $? I recently spend some time on cooking and I'm curious about the time evolution of the temperature of the water. I did some experiment and the temperature is of the form
$$ T = 100 \mathrm{^\circ C} + (T(t_0) - 100 \mathrm{^\circ C}) e^{-k(t - t_0)} $$
for some positive constant $k$ somehow (for example, see my previous post and my answer). In other words, 
$$ \dfrac{dT}{dt} = -k(T - 100 \mathrm{^\circ C}). $$
My question: How can I determine the coefficient $k$? Of course, I can determine it by experiments as shown below. But what I want to do is understanding this as a function of volume of water $V$, power of heater $P$, and some other variables related to a pot. So my expected answer is something like "If the volume (power) is $n$ times larger then the coefficient is $n$ times ($1/n$ times) larger."
Here is the experimental data and fitting graphs.
($V$ water is in the same pot with radius 9cm and heated by IH correspond to power $P$. I measure its temperature every 30 seconds.) 
My aim is determine the time evolution of the temperature of the water a priori. So if another approach is better, please tell me. I would appreciate if you help me. Thank you.
 A: The relevant law is conservation of energy.
$$ \text{change in thermal energy of the water} =  \text{energy put in} -\text{energy lost to air} $$
Apply Newton's law of cooling to write the energy lost to air.  Energy input from the eye is constant.  Then the change in energy of the water is a matter of notation I'm introducing.
$$ \frac{ d Q}{dt} = \dot{Q} = \dot{W} - A h \left( T(t) - T_{air} \right) $$
Also note that from definition of specific heat, the energy in the pot is $Q = C_p m T$, which is ignoring any offset, which isn't a problem because all temperatures can be relative.
$$ \frac{ d Q}{dt} = \frac{ dT}{dt} m C_p = \dot{W} + A h T_{air} - A h T(t)  $$
$$ \frac{ dT}{dt}  = \frac{\dot{W}}{m C_p} + \frac{ A h }{ m C_p} T_{air} - \frac{ A h }{m C_p} T(t)  $$
This is the form that I wanted, so that I could substitute in your $\beta$.
$$ \frac{ dT}{dt}  = \frac{\dot{W}}{m C_p} + \beta T_{air} - \beta T(t)  $$
The substitution I just did defines $\beta$.  Unit-wise, you can describe this more thoroughly.  But in short, $h$ is the area-specific heat transfer coefficient.
To create useful experiments, let's look at the steady-state solution of this equation.
$$ \beta T(\infty)  = \frac{\dot{W}}{m C_p} + \beta T_{air}  $$
I'll presume that you have a direct measure of the temperature of the air as well as the pot.  That leaves us with two unknowns.  We don't know $\beta$, and I don't agree that you know $\dot{W}$.  I'm forced to treat it as an unknown.  Actually, I would introduce another variable $\alpha =\frac{\dot{W}}{m C_p}$.
$$ \beta T(\infty)  = \alpha + \beta T_{air}  $$
$$ T(\infty)  = \frac{ \alpha }{ \beta} + T_{air}  $$
Even though we don't know $\alpha$, I might make use of the linearity that comes with it.  In other words, if you use 200 W and then 400 W, then I'm fairly sure the heat input increases by a factor of 2.  For $\beta$ we have to be careful.  You can't be (very) close to boiling or else it'll affect the convective currents.  And the water level will also affect its value a great deal (mostly due to $m$, but also $A$).
Experiment proposal
step 1
For any arbitrary water volume, get steady-state values for different power levels, and use this to ascertain $\alpha/\beta$, which is specific to that power level and volume.  Mathematically, that gives:
$$ \frac{ \alpha }{ \beta} = T(\infty) - T_{air} $$
You might want to do this for several power levels.  Just increase the power level a little, wait a little while, then record the temperature.  That should be linear, but the degree to which it's linear is a good test of how accurate the mental picture is.
step 2
We've been changing the power level, so now let's stop, keep both the volume of water and the power level the same, and look at the change in temperature over time.  The differential equation I wrote has a solution that basically comes out to the following, if the pot starts out at ambient temperature and then it heated.
$$ T(t) = T_{air} + \left( 1 - e^{ - \beta t } \right)  \left( T(\infty) - T_{air} \right) $$
You should know from the previous step what the final temperature is, so that's a known.
$$ \beta = - \ln{ \left( - \frac{ T(t) - T_{air} }{  T(\infty) - T_{air}  } + 1 \right) } / t $$
$$ \beta =  \ln{ \left( \frac{    T(\infty) - T_{air}  }{  T(\infty) - T(t)  }  \right) } / t $$
Provided you've correctly started at air temperature and that you correctly found the final temperature, you only need one data point in the middle to solve the above equation.
This gives a way to find time constant of heating, as well as ways to interpret the coefficients.  Remember that the time constant is specific to the amount of water in the pot.  If you wanted to get a more comprehensive understanding, I would use the data to solve for $\alpha$, and then correlate that to what the stove says is the heat, to get a correction factor of the amount of heat lost directly to air from the eye.
A: You have three processes going on:


*

*the heat injected heats the water at a constant rate

*the pan cools according to Newton's law i.e. the rate of heat loss is proportional to the temperature difference between the pan and the environment

*you lose heat due to water evaporation
It's easy to write down a differential equation for mechanisms 1 and 2, but I think mechanism 3 is going to be troublesome. There is some info about it on the Engineering Toolbox web site but this requires parameters like the effective moisture content of the air around the pan. I suspect you'll need to measure this rather than try and calculate it from first principles.
Note also that when the temperature reaches 100ºC the mechanism changes since the evaporation becomes effectively infinite i.e. even with a large energy input you can't raise the water temperature (much) above 100ºC. This means you'll get a discontinuity in the heating curve at 100ºC.
You can get some idea of the contribution from evaporative cooling (below 100ºC) by heating the pan with and without a tight fitting pan lid.
A: see we use simple integration ..
$$dT/dt = -k(T - 100)$$ 
this is in accordance with Newtons Law Of Cooling which says that the negative rate of change of temperature is directly proportional to the difference of the initial temperature of the substance and the surrounding temperature of the environment.
$$dT/dt = -k(T- 100)$$
$$dT/(T- 100) = -k \,dt$$
$$\int dT/(T- 100) = \int -k \,dt$$
$$\ln(T - 100) = -kt + C   \qquad \text{(1)}$$
Putting $t = 0$ at the time of origin  and the corresponding value of $T$, we get the value of $C$.
Putting the value of $C$ in (1) we can find $k$ in terms of $T$ and $t$. 
We can put the respective values for the final answer.
