# Thermodynamics at the stoves, cooking problem

I was wondering about a daily question with respect upon heat, time and thermodynamics "made easy".

Supposing I put a pot on the stove (we can say without losing so much that the pot is a cylindrical pot, with radios $R$ and being $h$, so the total volume is $V = \pi R h$). There is $1$ L of water inside it, namely $1$ kg of water (let's make it easy).

Now I turn on the fire, and I wait until the water is boiling. I split the problem in two different cases:

Case 1: Pot without the lid

In this case, I could state the following (if it holds): the amount of heat necessary for the water to be brought to the boiling point is

$$Q = m\mathcal{C}\Delta T$$

where $m = 1~\textrm{kg}_;\, \mathcal{C} = 4186~\mathrm{J/(kg\; K)}$ and $\Delta T = 80$ degrees just for simplicity, supposing I have to heat water from $20$ degrees to $100$ Celsius degrees (anyway the $\Delta$ will always be $80$).

I'm expecting then

$$Q = 334,880\ ~\mathrm J$$

of necessary energy, but the pot has no lid, so is that formula valid in this case? How can I determine the extra amount of required heat? This relates to the second case:

Case 2: pot with the lid

In this case, I do expect a $Q'$ such that

$$Q' < Q$$

because with the lid, there are no dissipation of energy out of the pot (nay: very small, I guess).

If also I take the time for the first and second experiment, I will surely state that

$$t' < t$$

Final questions

1) What could be the correct physical description of the two phenomena? Is there a different equation that has to be used when one treats those kinds of problems "with or without lids"?

2) Is there a way by which I can determine the necessary amount of time for the water to boil, knowing the volume of the pot, the material and maybe other parameters?

Here you are considering 2 different processes: in case 1 the process occurs at constant pressure (that of the atmosphere); in case 2 it occurs at constant volume (that of the pot). The coefficients that relate the heat exchanged in the process with the change in temperature are the (molar) heat capacity at constant pressure $c_p$ and constant volume $c_v$ respectively.

$$c_p = \frac{1}{N}\left( \frac{\mathrm{d} Q}{\,d T} \right)_p$$

$$c_V = \frac{1}{N}\left( \frac{\mathrm{d} Q}{\,d T} \right)_V$$

where $N$ is the number of moles in your system. (in ${\mathrm{d} Q}$ the differential is an "inexact" one, but I don't know how to type it here).

These quantities are related by: $$c_p = c_V + \frac{TV\alpha^2}{N\kappa_T}$$ where $\alpha$ is the thermal expansion coefficient and $\kappa_T$ the isothermal compressibility.

Therefore the heat exchanged in process 1 will in general be different than that exchanged in process 2 for the same $\Delta T$.

Finally you can approximately estimate the time needed by water to boil in the two processes. For example you can assume the all the work generated by the stove is converted to heat that will be absorbed by the water. That means that if you are using a stove with a power of $1500W$ the water in the pot will absorb $1500J$ of heat every second. From $\Delta Q = Nc \Delta T$ ($c= c_p \, \text{or } c_V$ according to the case you are considering) you can find the heat needed to increase the temperature of $\Delta T$ and hence the time. Of course different approximations are made here but this should give an idea.

• What if there is no air above the water in the constant volume pot, so that the water completely fills the pot to begin with? What do you think the boiling point of the water will be? – Chet Miller Feb 10 '16 at 16:44
• Short answer: yes it will start to boil, long answer is well described here: physics.stackexchange.com/questions/160114/… – NNec Feb 10 '16 at 19:53
• This reference is incorrect, starting with the sentence "As we add heat, the state will follow a constant pressure line to point 2." Actually, as we add heat, the temperature will increase, and, at constant volume, this will cause the pressure to increase. The pressure in the fluid will always be higher than the equilibrium vapor pressure at the existing temperature. So the fluid will never boil. – Chet Miller Feb 10 '16 at 20:05
• Yes it is true the reference consider the water can expand, I guess that if the container is really completely filled with water and it is perfectly smooth, such that nucleation of water bubbles on its surface is excluded, as the temperature is increased the water will become a supercritical fluid without any phase transition – NNec Feb 11 '16 at 19:01
• Another interesting situation is where there is just a small amount of head space and there is air in the head space (with the total initial pressure equal to 1 atm). With a small amount of head space, the vapor will be able to equilibrate very rapidly with the liquid. Then, with the air present, the total pressure will always be higher than the equilibrium vapor pressure of the liquid. Also, the air pressure will be rising with temperature. So the vapor pressure of the liquid will not be able to exceed the total pressure. – Chet Miller Feb 11 '16 at 19:46

The formula is valid, but you have to consider $Q$ to be the net amount of heat transferred into the water.

In both cases, as the water gets hot, it will radiate some heat to the environment (because the region above the water is cooler). In addition, the lidless pot will lose heat (and some mass) through evaporation.

The problem is that the amount of heat lost in these cases are not easy to model. That's why simple thermodynamics problems assume no heat loss. The evaporative heat loss is going to depend on the temperature of the water, the surface area of the water, the shape of the vessel above the water line, the humidity of the air, and other things like airflow/breezes.

1. The equation is fine, but (assuming minimal water loss) you're calculating net heat flow, not heat delivered from the stove.
2. Not easily from that data. It would probably be easier to do measurements of a few methods and then model the results.

I think that question 1 is pretty much answered by NNec. There is a difference on the amount of energy that your liquid would need to reach the required temperature due to the constant volume-constant pressure difference in heat capacity. But I am not sure about how "big" this amount is for you to need to take it into account. That is if the target is a specific temperature. If the target is for the water to boil it is a bit different, because at the lid-full case water would boil at a slightly higher temperature (due to higher pressure) than at the lid-less case (constant pressure, atmospheric).

HOWEVER, there is a difference between the amount of energy that a specific mass of water (1 kg, in your example) needs to reach your targeted temperature and the amount of energy that you have to give through the stove* due to energy dissipation, the amount of which (dissipated energy) differs a lot more in the lid-full and lid-less cases than heat capacity does. Your pot will lose some energy through the side surfaces and through the top. The latter is much much greater if there is no lid on your pot as heat is lost along with mass due to evaporation. Take a look at this site. I guess it might help you understand the heat loss concept in more detail.

$*$ Is it a stove or a hot plate? Because there is a huge difference between these two. Stove would heat up the whole environment around the pot and the water inside the pot in parallel so the energy dissipation through the site surfaces would probably not exist while the evaporation of the liquid would gradually be reduced due to environment saturation on steam molecules. But if you mean a hot plate then the energy dissipation would be much bigger.