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.

  • $\begingroup$ I'm trying to answer, but I'm worried there might be some iteration to this question. Firstly, what is the 100 degrees doing in there? You're using Newton's law of heating/cooling, and the two relevant temperatures are that of the water, and that of the air. Also, you'll need a new form of that equation which includes the additional heat input. A complication is that you can't really believe the number of Watts, because not all make it into the pot. That gives 2 independent variables, but we should be able to solve for them 1 by 1. $\endgroup$ – Alan Rominger Apr 5 '13 at 14:36
  • $\begingroup$ What AlanSE said and also: How did you determine $\beta$? Your numbers seem wrong. Larger $\beta$ corresponds to a faster temperature rise, but your numbers are going the other way. Also just from the look of your data you can't claim anywhere near three significant figures. :) Once you identify the important parameters you could probably get a reasonable estimate for $\beta$ just by dimensional analysis. $\endgroup$ – Michael Apr 5 '13 at 14:42
  • $\begingroup$ @AlanSE I thought the graph is linear. But if so, the temperature gets higher than 100$^\circ\mathrm{C}$. So I tried some other increasing function bounded above by 100. That's how I come up with it. Soon after, I rewrite it of form of the Newton's cooling law. For second part, I also think about it. But I'm unsure which heating values are primary: the one escaped from side via pot and the other escaped from the surface of the water. So I tried to solve simplest one but I got stuck. $\endgroup$ – Orat Apr 5 '13 at 14:49
  • $\begingroup$ @MichaelBrown I determine the $\beta$ with temperature of 0 seconds and 180 seconds. But it is likely that my experiments are wrong. I just do it with ordinary kitchen objects. It's not accurate. I already tried the dimensional analysis but the important parameters are unknown to me (I major in mathematics and I'm unfamiliar with these kind of problems). $\endgroup$ – Orat Apr 5 '13 at 14:54
  • $\begingroup$ I looked at your data and fits, and in my opinion, your formula is a rather bad fit (out of error bars in some places). Since the system is undergoing an obvious transition from one state to another (water starts to boil), maybe you could reach better fits if you'd try to split data to two domains (before boiling and in-boiling), and fit them independently, or with some matching condition at the border of domains. $\endgroup$ – firtree May 11 '13 at 11:31

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.

  • $\begingroup$ Since I know little about physics, it's hard to follow your explanation. I'll take time to understand it. Let me ask about the notation. In first equation, $W$ stands for work that heater do and hence $\dot{W} = 700\mathrm{W}, 1000\mathrm{W}$ in this case, right? And what $A$ and $h$ stands for? $\endgroup$ – Orat Apr 5 '13 at 15:35
  • $\begingroup$ @Taro Yes, you understand. The dots indicate that they are per unit time. Q and W and measures of heat, and dotting them makes them a measure of heat flow. A is area and h is the heat transfer coefficient, which you just have to Google for. None of these, however, are necessary for your problem. Mathematically, I made only 2 predictions that matter: the temperature goes linearly with the heat input and as a decreasing exponential with time. That means you have two parameters to solve for. I also should have written $T(\infty)=T_{air}+\gamma \dot{W}$ as an alternative to the heating coef $\endgroup$ – Alan Rominger Apr 5 '13 at 15:43
  • $\begingroup$ I get a little confused in notation $T(\infty)$. I think it make sense because the temperature is increasing and bounded above; hence there exists $T_\infty = \lim_{t \to \infty}T(t)$. So for sufficient large $t$, its derivative $dT/dt$ is zero and it leads to $\alpha/\beta = T_\infty - T_\text{air}$. However in this case, $T_\infty = 100^\circ\mathrm{C}$, isn't it? If so, $\alpha/\beta$ is only depend on only the room temperature? Is that means $h \propto T_\text{air}$? And I get $T = T_\text{air} + \dfrac{\alpha}{\beta} + e^{-\beta t}$. (Probably this might be just a little typo.) $\endgroup$ – Orat Apr 5 '13 at 16:37
  • $\begingroup$ @Taro The physics we have discussed so far are heat balance and Newton's law of cooling. If the final temperature will be 100 C, then I take it you mean that it will boil. That is extremely problematic. All of the assumptions we used are no longer valid. The behavior is piecewise (in the most simple sense). If you boil, you can't find $T_{\infty}$ in the same way. You could still apply these equations for heatup up until boiling, maybe by fitting an exponential fit to the data in Excel. You could do that with the data you posted. But with boiling onset the physics change. $\endgroup$ – Alan Rominger Apr 5 '13 at 17:02
  • $\begingroup$ I'm afraid I don't understand well. If the equations become invalid in boiling point, what do you mean by "steady-state solution"? Would you elaborate on it, please? $\endgroup$ – Orat Apr 6 '13 at 10:53

You have three processes going on:

  1. the heat injected heats the water at a constant rate

  2. 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

  3. 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.

  • $\begingroup$ I'm looking at your link, trying to find an occurrence of water temperature in the expression for evaporation rate, and it's hard. I'm thinking it's baked into the partial pressure of water vapor at saturation conditions. I guess that's probably not linear... $\endgroup$ – Alan Rominger Apr 5 '13 at 15:27
  • $\begingroup$ @AlanSE: yes, that's how I interpreted the info on that page. There are empirical fits for vapour pressure vs temperature, but it's all going to get a bit messy. $\endgroup$ – John Rennie Apr 5 '13 at 15:54
  • $\begingroup$ I recently solved a problem of my previous question and stated thinking about this problem again. At there, a term $T - 100\mathrm{^\circ C}$ appears. Your suggestion doesn't seem to contain this term. (And I cannot derive this term from any first principles at now.) What do you think about it? Is it an accident that the term fits? $\endgroup$ – Orat Apr 13 '13 at 13:26
  • $\begingroup$ The temperature can't rise above 100C because the water starts boiling, but I think there is a non-zero value of dT/dt just before you reach 100C and there is a discontinuity at 100C. In practice this would be very difficult to see because heat flow through the water takes a finite time so the water won't all be at the same temperature. My guess is that this washes out the discontinuity so an equation that approaches 100C asymptotically gives a good fit even though it's not strictly true. $\endgroup$ – John Rennie Apr 13 '13 at 14:07

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.

  • 1
    $\begingroup$ Thank you for your answer. But that's basically what I did in my previous post as I wrote in the question... The problem is how to determine the $k$ with respect to surroundings something like $V$ or $P$. And where the DE comes from if it really governs the system. $\endgroup$ – Orat May 11 '13 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.