Consider I want to heat 1L water in a 1L aluminum container. That container I put on my stove that I heat with gas. I want to know, what is the optimum energy I can deliver to cook that water efficiently? In my opinion, if I heat that water with too big fire (if I burn to many gas) maybe just some percentage of that heat will be absorbed by that water. Hence, the gas utilization will be not optimum. But vice versa, if I heat the water with too small fire, the water of course will be heated but the heat absorbed by the water will also be released to the environment (2nd Law of Thermodynamic) so efficiency of the gas of the stove's utilization will also be not optimum as the heat absorption of and heat release rate (HRR) maybe higher. For this question, just consider I can exactly control the gas flow rate to the stove so the heat it can release is controllable. Or in a more realistic, I can use electric jar heater which its power I can control exactly.

Before posting this question, I have searched many pertaining thermodynamics, such as specific heat capacity (c), heat capacity (C), Heat Release Rate (HRR), energy content of gas (in mega joule per liter), rate of heat flow (Q/delta-t), so did heat absorption rate (HAR) which is formulated as Q=mcdelta-T (which delta-T=the changing of temperature). But still I could not find the way to link between all those info to that my the above question: how to calculate the optimum heat absorption rate of an object?

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    $\begingroup$ You have to change the word substance to object as geometry plays an important role in heat transfer. $\endgroup$ Jan 1 at 8:39
  • $\begingroup$ Model it out with your geometric constraints. Ideally, one would deliver all energy instantaneously to where it would be in the end state. Can only heat the aluminum? That's a constraint but maybe your aluminum can get close to that ideal by becoming as distributed in the water as possible (think heat sink). How about as a manifold of tubes separating your fuel and air until they reach desired locations, causing combustion to occur closer to that ideal? $\endgroup$
    – Abel
    Jan 1 at 13:40
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    $\begingroup$ This will depend on the exact physical setup. Increasing insulation of the load (to the environment) would make the optimal output lower. Increasing the heat exchange efficiency from flame to load would make the optimal power output higher. $\endgroup$
    – Drew
    Jan 1 at 16:19
  • $\begingroup$ Evaluate the heat output of the gas to the heat input to the water plus losses. This should balance else you missed something. All heat transfer is by 3 ways : conduction, convection and radiation. $\endgroup$
    – Solar Mike
    Jan 1 at 23:25
  • $\begingroup$ @SolarMike, what is the formula to deal with absorption rate in/per unit of time? I.e., 1 calorie/(gram second) (Q/kg.t). I knew heat transfer rate which also Q/t, but it is related to the heat can released by a particular object/substance and doesn't tell about volume or its weight. For example, how much heat of a kg of water can absorb in a second, or something like so. I believe any substance need time to absorb it, it can not change the phase instantly. $\endgroup$ Jan 2 at 8:07

2 Answers 2


You need to quantify both competing factors. This can be done empirically without too much difficulty.

  1. Determine the rate of heat loss from the water to the environment vs temperature. Simply start with the water hot, and record the temperature over time with a temp sensor.
  2. Determine the heating efficiency from fuel to water vs output %. You could do this by weighting the fuel can before and after a run. Measure the temperature increase over time at different % outputs, cancel out the heat loss you measured previously.

Number 2 doubles as an actual empirical test, since you will end up with a table of efficiency values for each output %. But it also will allow you to write the equations needed to determine the ideal % analytically.

You could write either of these equations using thermodynamics only, but number 2 would be extremely difficult, and unlikely to be accurate without computer simulations.

  • $\begingroup$ Of course I can do experiment. But I need formula, which I believe there is, to link between the supplied heat to the released heat. Regarding the number two, as I have said, rather than using gas stove I can use electric heater jar so it is exactly measurable. $\endgroup$ Jan 1 at 18:06
  • $\begingroup$ Each experiment will produce a graph. You can best fit that graph to produce an equation relating the 2 factors. The ideal thermodynamic equations to figure out number 2 would be complicated and likely inaccurate. $\endgroup$
    – Drew
    Jan 1 at 19:36

You face a complex problem and have three likely choices for the approach to solve it. The first is to do experiments on what you define as your measure of effectiveness versus flame size. Perhaps you define effectiveness as the time to heat the water to a certain temperature. Shorter times mean greater effectiveness. The second choice is to model the system using computational tools such as COMSOL. The third is to establish an effective control volume approach to analytical analysis. You will need to do take steps on the third option to do the second option. So, what might be the starting point for an effective control volume approach? It will be to draw a system picture that defines all geometric and materials parameters needed to model the system. Here is a start on the concepts.

The heat from the burner will go to two places, the container and the external surroundings. You increase the amount of heat going to both by turning up the flame. Calling each heat rate $\dot{q}$, energy conservation gives

$$\dot{q}_{flame} = \dot{q}_{con} + \dot{q}_{surr}$$

The ratio $\dot{q}_{con}/\dot{q}_{surr}$ depends on the geometry of the flame, such as the size of the flame and the distance between the flame to pot. The ideal case is to have this ratio be as large as possible. The amount of heat lost to the surroundings is due to convection and radiation, modeled simply as below with respective convection coefficient $h$, area $A$, emissivity $\epsilon$, Stefan-Boltzmann constant $\sigma$, view factor $F$, and temperatures $T_j$.

$$\dot{q}_{surr} = h_{air}\ A_{surr}\ (T_{flame} - T_{surr}) + \epsilon_{air}\ \sigma\ F\ A_{surr}\ (T_{flame}^4 - T_{surr}^4)$$

The heat going into the container will go to the pot and the water. The rate that each is heated depends on geometry factors such as the sizes (diameter and heat) of the pot and the water (volume) as well as the materials properties, such as thermal conductivities and heat capacities. In energy conservation, we write

$$\dot{q}_{con} = \dot{q}_{pot} + \dot{q}_{w}$$

In non-steady state (NSS) analysis, the heat inputs will change the temperatures for the pot and water as a function of time. The NSS analysis requires factors such as the Biot number, will lead to a second order differential equation $d^2T/dt^2$ for temperature change with time, and could eventually lead to complex deconstructions such as Hessler charts. At steady state (SS), we might simply say the system is lumped with the pot and water at the same constant temperature.

During both the NSS and SS processes, heat going into the pot and water will dissipate to the surrounding air. The rate of dissipation depends on geometry factors such as the contact area to volume ratio of the pot or water versus the surrounding air as well as thermal conductivities and convection coefficients of the air, water, and pot. The NSS analysis has a comparable approach to above with the surroundings, only we can reasonably neglect radiation. The losses are strictly convection from the pot and the water surface areas. In a pseudo-lumped approach, we might write

$$\dot{q}_{con} = h_{air}\ A_{con}\ (T_{con}(t) - T_{air}) $$

From the above, you see that even in the simplest case for a pseudo-lumped analysis, you will need such basic factors as the flame, surroundings, and container temperatures; the various areas for flame to surroundings as well as container to surroundings; and the air convection coefficient. Any desire to do analysis for NSS will demand more effort, include truer definitions for the component physical geometries (volume to area ratios) as well as additional materials properties (thermal conductivities).

Hope this get you started.


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