I Never Took Thermodynamics...

[chinbeard]

Liquid 1 is heated and then allowed to cool to room temperature. Liquid 2, which is at room temperature, is added to Liquid 1 as it cools. As long as Liquid 1 is still hotter than room temperature, this will bring down the absolute temperature. If Liquid 2 is added earlier (when Liquid 1 is hotter), would it bring about a greater change than if it were added later (when Liquid 1 is closer to room temperature)?

[D'Art]

It depends on what the two liquids are.

[tchwojko]

Assuming the same volume of liquid 2 is added in either case, then I think the answer is yes, if liquid 2 transfers heat better than the air. (The temperature of the container of liquid 1 also matters.)

It's been a long while since my college Thermogoddammits class.

[Hauptman]

Liquid 1 is heated and then allowed to cool to room temperature. Liquid 2, which is at room temperature, is added to Liquid 1 as it cools. As long as Liquid 1 is still hotter than room temperature, this will bring down the absolute temperature. If Liquid 2 is added earlier (when Liquid 1 is hotter), would it bring about a greater change than if it were added later (when Liquid 1 is closer to room temperature)?

On it's face, it would seem the answer is yes.

The temperature of liquid 2 is contant in both examples; room temperature. The only variable is the temp of liquid 1.

If the temp of liquid 1 is higher then there is more heat to transfer to liquid 2 when the two are combined, so the change would be greater than if liquid 1 starts at a lower temp.

Or you could say that when the difference in temperatures between the liquids is greater then you will have a greater change in temperature when they are combined.

[keith]

Liquid 1 is heated and then allowed to cool to room temperature. Liquid 2, which is at room temperature, is added to Liquid 1 as it cools. As long as Liquid 1 is still hotter than room temperature, this will bring down the absolute temperature. If Liquid 2 is added earlier (when Liquid 1 is hotter), would it bring about a greater change than if it were added later (when Liquid 1 is closer to room temperature)?

Which liquid is extending it's arm?



Agree with the above unless the question is phrased sneakily. There are after all situations where the numerical change is greater but the % change is the same.

[chinbeard]

This happened IRL the other night. Liquid 1 was a big pot of lentil soup. Liquid 2 was a 28 oz. can of diced tomatoes. Since the tomatoes didn't really need cooking, and since the big pot of soup needed to be cooled (I didn't want to run up my energy bill by sticking a big pot of molten vegetables in the refrigerator), I added the tomatoes while it was still hot.

In this case, there was so much hot soup that the tomatoes didn't make a noticeable impact. This made me wonder if the timing really mattered. I figured that it must, since no change would happen if I waited until Liquid 1 reached room temperature.

[Hauptman]

This happened IRL the other night. Liquid 1 was a big pot of lentil soup. Liquid 2 was a 28 oz. can of diced tomatoes. Since the tomatoes didn't really need cooking, and since the big pot of soup needed to be cooled (I didn't want to run up my energy bill by sticking a big pot of molten vegetables in the refrigerator), I added the tomatoes while it was still hot.

In this case, there was so much hot soup that the tomatoes didn't make a noticeable impact. This made me wonder if the timing really mattered. I figured that it must, since no change would happen if I waited until Liquid 1 reached room temperature.

I understand now.

The phrasing of the question is tricky; you asked about the change in temperature when in fact you're looking for the change in time for reaching a desired temperature.

So from your example, if your desire is to get the hot soup down to room temperature then it doesn't matter when you add the tomatoes. In the end it will still take the same amount of time.

If the soup it hotter when you add the tomatoes then the change in temperature will be greater, but the resulting temperature of the soup overall will be higher at the moment you combine them than if you add the tomatoes when the soup is at a lower temperature. After you combine them you still have to wait for other factors to cool the soup.

Of course, this all assumes that all other things are equal when in reality they're not.

The cooling of the soup is mainly dependent on the air circulation around the container. The air circulation is affected by the surface area of the soup (and it's container). When you add the tomatoes you are potentially adding surface area which will affect the cooling rate, but the volume of the two liquids and the shape of the container will also affect the surface area, so without more information it's impossible to know the significance of these factors.

Doesn't science make things so much more understandable?

[fencerchica]

When I first saw this thread I thought it was a homebrewing scenario! For me this would be a question about whether it's better, when doing extract + mini mash style brewing, to cool your small volume of boiling wort first and then add room temperature water to bring it up to desired volume, or to add the room temperature water first in hopes of cooling down the boiling wort a little faster.

(For me, the answer is that cooling the hot wort first makes more sense because in that case you can pack the vessel in ice or use a heat exchanger, whereas the larger volume after adding the room temperature water is much more cumbersome to deal with.)

[PretAllez]

Actually, you are adding a liquid and a solid to your hot liquid (which also has solids (lentils), so you should factor in the energy absorbed by the tomato cooking process (which will depend on the surface area in contact with the soup (were they chopped or whole?) and, possibly, acidity level, which might change the heat-absorption properties of your lentils).
Now, given that tomatoes are mostly water, this should not affect your time-to-cool calculations too much, but if you were adding potatoes, well that would be a whole different kettle of soup. Potatoes absorb a great deal of energy in cooking, and would cool your soup down much more quickly than tomatoes (but you might end up with more of a stew...)

(My grandfather was a chef in rural France and his fail-safe remedy for a chimney fire was to throw in some cut, raw potatoes, which absorb so much energy they effectively put out the fire)

[Rabid Monk]

Liquid 1 is heated and then allowed to cool to room temperature. Liquid 2, which is at room temperature, is added to Liquid 1 as it cools. As long as Liquid 1 is still hotter than room temperature, this will bring down the absolute temperature. If Liquid 2 is added earlier (when Liquid 1 is hotter), would it bring about a greater change than if it were added later (when Liquid 1 is closer to room temperature)?

It's been awhile, and my mouse just stopped working, so bear with me.

I'm also going to define my initial conditions as being immediately before the room-temperature liquid is introduced to the above-room-temperature liquid.



The question:

Liquid 1 (L1), at a temperature (Ti) greater than room temperature (Ti > Ta, where "a" means ambient), and is cooling via free convection.

Liquid 2 (L2), initially at room temperature (Ta), is introduced to L1.

Does the resultant change in temperature of the mixture (Ti - Tm = dT) depend on the value of Ti? That is, does a greater Ti result in a greater dT?

The point of interest here to determine if the L1 can be cooled faster by introducing the room-temperature L2 into the mix earlier.


List of Terms:

L1: liquid one, initially at Ti
L2: liquid two, initially at Ta
Ti: Initial temperature of L1, Ti1 > Ta
Ta: ambient temperature
Tm: resultant temperature of mixture
Th: hot temperature
Tl: cold temperature
dT: change in temperature
V1: volume of L1
V2: volume of L2
dE: change in Energy
Q: heat transferred to system
W: work done
m: mass
cp: specific heat at constant pressure (not to be confused with cv, which is specific heat at constant volume, although the two terms are identical if the fluid is incompressible)


Assumptions:

- "Instantaneous mixing", i.e. No convection (or radiation) occurs between the introduction of L2 and full mix (either between L1 and the surroundings, or between L1 and L2). I make this assumption because otherwise I would have to make assumptions about surface area, atmospheric conditions, and a few other variables which I don't have the information for. It also means my energy balance has less terms to deal with.
- No heat is being added to the heated liquid. This is not explicitly stated in the question (the cooling rate could be being controlled), but it is implied. So, to make it official, i am assuming that L1 has been removed from the heat source(s).


System Description:

We are going to take our control volume (the space in which the heat transfer takes place) as an imaginary container, perfectly insulated (no heat transfer to surroundings), with both L1 and L2 in it, separated by an impermeable barrier (no fluid transfer) which is also a perfect insulator.

At the time of mixing, the barrier vanishes. L1 and L2 mix perfectly and instantly.


Basic Equations:

At its simplest, this is a simple energy balance problem.

Having assumed that there is not heat transfer between the liquids and their surroundings, we only have to look at a control volume originally containing the hot liquid (or the cold one, your choice) and the same control volume after mixing, and find out the change. Our basic equation (indeed, the basic equation for all thermodynamics, is:

dE = Q – W

Since we are assuming the mixing is instantaneous, there is no work done. Work is a function of time. So we get:

dE = Q

Q, in turn, is expressed as:

Q = mcp(high temperature – low temperature)
(This is simplified. It works because low – high = negative)

The heat transfer, then (Q), is the amount of heat gained or lost by the system. For the two liquids, this equation becomes:

Liquid 1: Q1 = m1 cp1 (Ti – Tm)
Liquid 2: Q2 = m2 cp2 (Ta - Tm)

Since both L1 and L2 are contained within our system, and are the only things in it, it follows that the change in heat energy for one will be equal to the change in heat energy for the other. That is:

Q1 = Q2

thus

m1 cp1 (Ti – Tm) = m2 cp2 (Tm – Ta)

Using this, we can solve for Tm. I’ll leave the algebra as homework for those who want to do it.
To make things simpler, I’ve defined the term m1/m2 as A, and cp1/cp2 as B.

Tm = (ABTi + Ta) / (AB + 1)

Note that if the two liquids are the same mass, A =1
Note that if the two liquids are the substance, B =1
If both L1 and L2 are the same substance, and both are of the same volume (densities being identical), then Tm = (Ti - Ta) / 2, as you would expect.

Recall from earlier that dT = Ti – Tm
We can now solve this, algebraically. Once again, the algebra is left up to the the masochists.

dT = Ti – Tm = (Ti - Ta) / (AB + 1)


Conclusion:

Our final formula for determining the change in temperature of the mixture is as follows:

dT = (Ti - Ta) / (AB + 1)

where
dT = temperature change
A = m1/m2
m1 = mass of liquid 1
m2 = mass of liquid 2
B = cp1/cp2
cp1 = specific heat at constant volume of liquid 1
cp2 = specific heat at constant volume of liquid 2
Ti = initial temperature of liquid 1
Ta = initial temperature of liquid 2, same as ambient

As you can plainly see, the initial temperature of liquid 1 does affect the heat lost.

The higher the temperature of liquid 1 when you mix in liquid 2, the greater the change in temperature. This is because there is a greater initial difference between the two.

However, we also know that from Tm = (ABTi + Ta) / (AB + 1), a greater initial temperature will also result in Tm being hotter.

If you're looking to reduce cooling time, you want the biggest change in temperature you can get. Put the room-temperature water in early. Be careful, watch out for splashes, and DON'T SCALD YOURSELF!!!

You could also lower the initial temperature of L2. It's winter, so throw it outside for a few minutes. Or in the fridge. Or even cold water from the tap.


Final Notes:

Just for those who are not strong in math, the mass of a liquid can be found as follows:

density * volume

Finding the density of a liquid is a matter of seconds on Google, so I'm not going to bother listing them. Especially as I suspect you are thinking of L1 and L2 being the same liquid. In which case, A = (volume of L1) / (volume of L2)

Ditto for volume. The formulas for that are easily available, and for pretty much any geometric shape you can imagine.

Finding the specific heat is also very easy to do online. Once again, use the Google. It knows all. But beware: If the liquid(s) you are talking about are not incompressible, cp and cv will not be the same. Be sure you select the correct one!

Finally, if this is too complicated for you, you can just go here, which uses the same equations I just derived. Albeit with slightly different terminology.




ETA: Finally posted and formatting cleaned up, and I must say, I think I might have learned something in engineering!

[Rabid Monk]

Actually, you are adding a liquid and a solid to your hot liquid (which also has solids (lentils), so you should factor in the energy absorbed by the tomato cooking process (which will depend on the surface area in contact with the soup (were they chopped or whole?) and, possibly, acidity level, which might change the heat-absorption properties of your lentils).
Now, given that tomatoes are mostly water, this should not affect your time-to-cool calculations too much, but if you were adding potatoes, well that would be a whole different kettle of soup. Potatoes absorb a great deal of energy in cooking, and would cool your soup down much more quickly than tomatoes (but you might end up with more of a stew...)

(My grandfather was a chef in rural France and his fail-safe remedy for a chimney fire was to throw in some cut, raw potatoes, which absorb so much energy they effectively put out the fire)

How big are the tomatoes?
You can always do a Lumped Capacitance analysis, if they're small enough...

Also, if you ever want to skin a tomato, dunk it quickly in boiling water. And then into cold water, unless you like cooked tomato (it grows on you).

[keith]

Also, if you ever want to skin a tomato, dunk it quickly in boiling water. And then into cold water, unless you like cooked tomato (it grows on you).

Remember to cut the skin first, makes the peeling quicker.