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So I'm trying to come up with a simple hack to heat up food in a water bath to hopefully +-1ºC of a target temperature, but I really don't understand the physics of it all so much. I'm pretty sure I can keep a hot plate at a somewhat steady temperature by cycling it on an off at regular intervals, but I'm not sure what that means for the food itself.

Here's what I'm thinking: If I can get the hot plate to stay steady at a temperature of my choice, when will place a pot of water on it and the water will stop heating up after it hits some lower temperature point, say at 75ºC. That means it has reached thermal equilibrium and is heating up at the same rate as it loses heat to the environment - and as long as the heat output of the plate is the same and the environment is stable, the water shouldn't drastically change temperature.

What I want to know is: will the final equilibrium temperature will be significantly different after putting some food in it?

I did some googling and managed to come with the following, probably wildly incorrect reasoning:

Most foods' specific heat is around 3 kJ/KgºC, and waters' is around 4 kJ/KgºC. I suppose that means given the same heat input, food will heat up about 1.33x more than water?

And thus a pot with, say, 3kg water and 1kg of meat (assuming a specific heat of 3 kJ/KgºC for the meat), all the stuff in the pot will have an average specific heat of 3.25 kJ/KgºC (3+3+3+4=13, divided by 4kg = 3.25), so it will all heat up 1.08x more?

What exactly does heating up 1.08x more mean, though? Do I just multiply 75ºC by 1.08? Surely not, right? Do I need to stick 1.08 in some formula or convert it to kelvin or something? What even are numbers, really? Thanks!

Just to clarify, what I'm concerned here is the final temperature once the pot reaches equilibrium again, not how much the temperature will drop when the food is added.

EDIT: To further clarify - what I want to calculate is what the final equilibrium temperature of a pot of water will be after placing some food in it, knowing only the final equilibrium temperature of the water before placing the food.

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  • $\begingroup$ If the hot plate keeps adding heat then the whole system will equalize at the 75 deg C. If however, you start with the water and container at 75 deg C and there is no more heat input then the total mass temperature will equalize somewhere below 70 deg C based on the mass of food to raise from its starting temperature, does it start at ambient or from the fridge or freezer? $\endgroup$
    – Solar Mike
    Apr 6 at 17:08
  • $\begingroup$ The hot plate would keep adding heat in my hypothetical scenario, but it sounds counterintuitive to me that the temperature would equalize at the same 75C. How can the same amount of heat applied to more mass reach the same temperature? $\endgroup$ Apr 6 at 17:10
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    $\begingroup$ You said you have set the hotplate to add heat to 75 deg C - what else do you expect or are you confusing heat and temperature? $\endgroup$
    – Solar Mike
    Apr 6 at 17:12
  • $\begingroup$ Maybe I am, I don't really have a physics background which is probably why I'm so confused about this. I meant in the example that I'd have set the hot plate at setting that made X amount water stay at a steady 75C. I imagined that adding stuff to the water would make the final temperature lower, since it is the same heat input affecting more mass? $\endgroup$ Apr 6 at 17:16
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At equilibrium, the heat entering the pot and the heat leaving the pot (heat flux levels) will be equal, otherwise the pot will either heat up or cool down.

What matters here isn't the heat capacity of the contents, it's the heat flux leaving the pot. I'm assuming that the heat from the hot plate will be constant in both scenarios. It's not clear to me exactly how the exiting heat flux changes with added materials (meaning what the equations are), but intuitively if the water level in the pot goes up, you should have more heat leaving it. Eventually, your burner wouldn't register hardly any temperature rise in the pot at all if the heated water had a very large surface area to transfer heat from. At the smaller water levels here, this would be the same really as just adding 7% more water to the pot (again we really need density of the items, water absorption, etc, but I'm discussing generalities). I would expect the heat loss change to be based proportional to the change of the height of the water column, so it would change less than the mass added to it. Change in height = volume added / (pi*r^2). So the change in temperature (in degrees K) would be proportional to the inverse of the square of the added volume.

I will assume for this that we are adding a significantly smaller amount of food than the amount of water, so that the heat transfer through the materials inside are immaterial. If you added 2 kg of carrots to a liter of water, the heat transfer through the materials will be less due to the loss of convective currents in the water. Add .2 kg and I doubt you would see the difference.

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  • $\begingroup$ I was thinking of using a covered pot for this application, so it wouldn't lose so much heat, and I imagine the main difference for the final result would be the added mass (and the heat capacity of that mass, but apparently that is not so relevant) $\endgroup$ Apr 6 at 16:53
  • $\begingroup$ but a higher water column will transfer more heat through the wall of the pot. $\endgroup$
    – Tiger Guy
    Apr 6 at 17:14
  • $\begingroup$ My apologies - I don't think I fully understood your post before, but now realize it perfectly answers my question: in this scenario, the temperature would stay basically the same. $\endgroup$ Apr 6 at 23:24
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TL;DR: The temperature in pot of water when food is placed in it is not only a matter of heat capacity. Other parameters like heat conductivity and convection have a significant effect on the transient temperature response.

I will try to answer by taking a different simpler example, before tackling your example. Before I do that, I will point out that water has probably the highest heat capacity out of common substances (at around 4 [kJ/kgK]). The reason that meat has approximately 3 [kJ/kgK] is because meat is about 70-75% water.

A short explanation of heat capacity

Heat capacity $c_p$ is the amount of heat energy in J you have to give to a kg of substance in order for it to raise 1 degree K. So in order to raise about 10 degrees Kelvin 1 kg of water you'd have to give it about 40 kJ. Equivalently, if you add 40 kJ of energy to 10 kg of water the temperature will increase only by 1 degree.

(Another quantity you might want to learn about is enthalpy, but for simplicity's sake we'll not be discussing about it).

Example:

Let assume that you have 1 kg of water at 20 $^oC$, and 1 kg of water at 60 $^oC$. If you put the two together in a pot (and assume that heat capacity remains constant - although it changes with temperature), then the temperature will be approximately 40 degrees.

So what happens is that the hotter water loses about 80 [kJ] $= (1[kg]\cdot 4\left [\frac{kJ}{kg K}\right]\cdot (-20) [K] $, while the cooler water gains those 80 [kJ] and raises its temperature by 20 [K]. (This is actually clearer when talking in terms of enthalpy but for this purpose this should be sufficient).

In a similar example, if there were 3 kg of 20$^o$C, and 1 kg at 60$^o$C, then the hotter will loose 120 kJ, which will be gained by the cooler water.

However, all the above are due to the fact that the quantities discussed above are liquids, and their instant mixture allows quick exchange of heat. Things are different for solids. The reason is that they are at a different phase.

different phase

Assume now that you have a ice cube of 1 kg, which is at -10 degrees, and 1kg water that is at approximately 20 degrees.

The heat energy in the system is equal to 2kg of water at 5$^oC$. However if you place the icecube in the water it will take some time until ice changes to water (if you could somehow put this into a thermally insulated container (with minimal heat capacity), the resulting temperature after a while will be 5$^oC$. However, for a long time you will have an icecube of -10$^oC$ with an outer layer which is close to zero.

Now, if your take an icepick and break up the ice cube and then place it into a blender, and then join them together, then almost instantaneously the result will be 2 kg of water at 5$^oC$.

heat conductivity k

So the idea that creeps up now is heat conductivity. This is the rate by which heat flows through a material. (Actually, its also convection, but this post is starting to look like a few weeks worth of lecture notes for a physics class, so I won't delve into that).

So heat conductivity is one of the main factors that affect the heat flux. That is why when you hold a metal spoon (high conductivity) over a fire, eventually you feel the temperature rise, while, if you hold a wooden spoon (low conductivity) you don't feel any change in temperature unless your hand is on fire. (Ice by the way has a surprisingly low heat conductivity coefficient which is one of the reason's why igloos are made of ice).

heat transfer mechanisms

The following image shows the different mechanisms of heat transfer. That might serve for the following (short) discussion of your example.

enter image description here

your example

Putting a piece of meat into water is subject to all the above (and more). When you put a piece of meat at 5$^oC$, into water of 45$^oC$, the heat isn't immediately exchanged. It takes some time for the heat to propagate into the meat. this is why you get the different zones in the cooked meat below (although this is not boiled).

enter image description here

Bottom line: The way heat transfers in a pot of water when food is added, is affected by many parameters (conductivity, convection and heat capacity). Maintaining a constant temperature in the water between $\pm 1^oC $ will be a challenge.

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  • $\begingroup$ Thanks! This gives me a much more thorough understanding of the thermodynamics in this scenario. That being said, I have learned by asking this elsewhere that for the specific application I mentioned (determining equilibrium or the steady-state solution), the defining factors at play are heat input and output - so I can disregard the added mass, making the entire premise of the question invalid - the final temperature of the contents of the pot would end up precisely the same! $\endgroup$ Apr 6 at 23:19
  • $\begingroup$ I apologize for not accepting your answer, since it is very informative and well-written - but @Tiger Guy's answer (which I just now understood) does answer my question more directly: The difference in temperature would be negligible in this scenario. $\endgroup$ Apr 6 at 23:27
  • $\begingroup$ That's ok. From your questions I could see that you were a bit confused about the concepts, so I opted for going back to the basics, while trying to explain why how does the temperature change and why its dynamic.. At least you found it informative. $\endgroup$
    – NMech
    Apr 6 at 23:57

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