# How much energy is required to boil and maintain boiling water temperature for a set time period?

I am trying to calculate the amount of energy required to bring 1 litre of water up to boil (from 20 degrees C to 100 degrees C) and to maintain the water temperature at 100 degrees C for a period of 5 mins.

The end goal is to translate this into the theoretical amount of wood required to do so over an open fire using the default net calorific value of wood (0.0156 Tj/ton), and assumed 10% efficiency of an open fire.

So far - using the Specific Heat of Water as 4.186 kJ/L I have calculated that the energy required to bring the water up to the boil would be:

1 x 4.186 x (100-20) = 334.88 kJ

My question is: how do you calculate the amount of energy required to maintain a water temperature of 100 degrees C over a period of 5 mins.

I have looked into using the Latent Heat of Water Evaporation value of 2260kJ/L multiplied by the % of water that is vaporised over that period. However- I do not have a figure for rate of vaporisation.

Secondary question: how do you calculate the rate of vaporisation assuming the power being applied by the heat source is constant?

• You may want to use Newton's law of cooling to calculate heat lost to the environment. – Sam Jul 3 '20 at 14:06
• Are you boiling water in an open container? If so, what are the dimensions of the container? How full is the container? What is the ambient temperature? Is the wind blowing? Your question may be difficult to answer, but it's not possible to answer without somewhat more information. – David White Jul 3 '20 at 16:03

Good Evening,

as every system you can write down an energy balance to compute the amount of heat that has to be given to keep your water at constant temperature.

$$\dot{Q}_{fire} = \dot{Q}_{losses}$$

Now your energy losses are represented by the energy transported by the vapour that flows out of your container and heat exchanges with the environment:

$$\dot{Q}_{losses} = \dot{m}_{vap} l_h + hA(T_w - T_{env})$$

$$\dot{m}_{vap}$$ being the amout of vapour escaping from the system (kg/s), $$l_h$$ the water latent heat. For the second term, $$h$$ (W/m2K) may be some equivalent of an exchange coefficient. The exchange takes place at the surface ($$A$$) between the water and the surroundings. In the end, $$T_w = 373$$ K and $$T_{env}$$ could be the temperature of the surrounding environment. Now $$h$$ is quite problematic to calculate, because it depends on a lot of factors that would make this explanation quite tedious.You can reasonably assume $$h = 50$$ W/m2k even though it may vary between 10 and 100.

• Many thanks @iterrate- this was very helpful. I wondered if you could provide some more insight on the exchange coefficient as some sources claim that for boiling water the heat transfer coefficient ranges between 3,000-100,000 W/m2K – EmmaDonn Jul 10 '20 at 9:45

Lets look at the units for power: J/s= (kJ/L)(L/s)(1000 J/kJ)

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Charles Francis Jul 3 '20 at 16:56
• He did ask how to relate the vaporization to the power. – R.W. Bird Jul 3 '20 at 18:48

Of course I can get in to the details of the exchange coefficient. I would say that the order of magnitude you find out are those typically found for water vapour in relatively "extreme" conditions such as those you can find in electric generation power plants. I was thinking more about the liquid water itself. The exchange coefficient is based on several properties, such as radiation, the fluid motion (convection) and thermal conductivities of everything is in contact with the water (container, air). It also depends on the geometry of the containter. Each of these mechanisms are object of University courses that I can never summarize here in a simple post. However I think there is a more simple and intuitive way to find out what could be a reasonable $$h$$. Imagine that you stop the fire at a given instant $$t_0$$. From that moment the temperature of the water would change following this simple differential equation: $$mc_p\frac{\partial T}{\partial t} = hA(T_{env} - T)$$. We neglect the loss of vapour for the moment, since they do not change the reasoning. The solution we search is in the form of $$T(t) = T_{env} + (T(t_0) - T_{env})e^{-\frac{hA}{mc_p} t}$$. At this stage you start to ask yourself how much could it take for the water to cool down to ambient temperature? You can do a guess or perform an experiment. In any case, you put your estimated $$t$$, togheter with the well known $$m$$ and $$c_p$$ and solve for $$h$$. I let you do the math. Let me know if $$h \approx 10$$ or $$\approx 10^4$$ !.

Regards