Tea kettle question - part two

[apophenia]

Thank you Rum, Mim, and Zilla. I love this kind of stuff. I was an avid reader of Gardner's mathematical games in SciAm back in the day.

Speaking of which, what is the answer to this?



(I knew back when I was a youngun, but I haven't been that in many a moon.)

"I'm just a poor old woman, my sight is dim, my eyes are poor, my nose is nackered."

[Warren Dew]

Speaking of which, what is the answer to this?

The red diagonal and the green diagonal are not in line, as 2:5 and 3:8 are not the same ratio.

[Rum]

You theoretical guys are useless. Do the experiment. First, define the temperature. Where in the cup?

I think the theory stuff is brilliant. I watched Brian Cox talking the other days about how some pure maths guys had worked out the upper possible density of atoms in a given area (or something like that). They worked this out simply using maths and no observations. They reached an upper limit of I think it was 1.4 times the mass of the Earth for any given concentration of packed atoms.

Observation then showed as predicted purely by abstract calculation that all white dwarf stars have an upper limit of 1.4 times the mass of the earth and conformed exactly to the abstract predictions.

I have probably got that totally arse backwards, but I was impressed none the less.

[Svartalf]

and anyway, actually experimenting would be wasteful (if simultaneous), and/or ruin the enjoyment from many a cuppa (if you have to tinker with several cups of beverage, minding protocol and measuring temperature over time)

[Gawdzilla Sama]

It's your own fault if the tea is too hot.

[Don't Panic]

Suppose you have a hot cup of coffee (or tea!) an you want it to cool down to make it drinkable. Is it quicker to leave it standing and then add room temperature milk when the liquid has cooled a bit or add the milk straight away? Which gets it to a drinkable (i.e. cooler) state quicker, if either?

Adding the milk immediately will cool it to a drinkable state faster, the bigger the temperature difference between the tea and the milk, the more energy required to equalise the temps, so the tea cools faster.

[Feck]

Speaking of which, what is the answer to this?

The red diagonal and the green diagonal are not in line, as 2:5 and 3:8 are not the same ratio.

no !

[PsychoSerenity]

If adding the milk doesn't cool it down enough by itself, I think it would take the same time for the whole mixture to reach the desired temperature, whether the milk is added at the beginning, or at precisely the time that the milk will cool it down to the desired temperature.

Nope. Adding the milk last is faster. The heat transfer from the liquid to the surrounding materials is more efficient the higher the temperature difference.

I thought I was already taking that into account. Doesn't the temperature drop, caused by adding the milk, also depend on the difference between the milk and the tea? So as it gets cooler, adding the milk won't cool it down so much, so you have to wait longer anyway, and it balances out?

I've tried to work it out with numbers, but I seem to have forgotten how they work.

No.

You can try to think at it like this. Your original combination of tea and milk has a certain amount of excess heat energy. This total energy does not change when you mix the liquids. So your task is to make the transfer of heat to the surroundings as efficient as possible.

Of course you could make an interesting case out of it if you took the milk out of the fridge at the same time as you draw your teacup. Then it would all depend on weather the milk heats faster or slower than the tea cools off. That could become a nice calculation for Jim's students

I've been doing some revision and working out how to do the sums, and I think you might be wrong. As far as I can see the amount the temperature goes down when you add the milk does also depend on the difference in temperature between the milk and the tea. And as the milk is at room temperature, it balances out. I'm not sure what would happen if the milk was at a lower temperature again... I'd have to do more sums for that.

Here's what I've worked through so far:

Initial Temp of Tea; T1 = 100
Ambient Temp and Temp of Milk; T2 = 10
Volume of Tea; V1= 4
Volume of Milk; V2 = 1
Rate of cooling; k = 1% per unit of time = 0.01
Ideal Temperature = 50
Temperature mixed; Tm
Time; t

First, calculate temperature, Tm, if milk added immediately.

Tm = (V1 * T1 + V2 * T2) / V1 + V2
Tm = (400 + 10) / 5
Tm = 82

Then calculate time until cooled to 50.

T(t) = Temperature at time t

Solution for Newton's Law of Cooling:
T(t) = T2 + (Tm - T2) e -kt

Solve for t, when T(t) = 50

50 = 10 + (82 - 10) e -0.01t
40/72 = e -0.01t
ln(40/72) = -0.01t
t = (ln(40/72))/-0.01
t = 58.78

Now we try it the other way, with the milk being added at the last minuet.

We need to work backwards from the ideal temperature, to find out the temperature of the tea when the milk needs to be added - Tx.

Tm = (V1 * Tx + V2 * T2) / V1 + V2
50 = (4 * Tx + 1 * 10) / 5
250 = 4Tx + 10
Tx = 60

Note that the temperature drop from adding the milk at 60 is only 10 degrees to 50, whereas when added at 100 it was 18 degrees to 82.

Tx = 60, becomes our new Ideal temperature, so we calculate the time it takes to cool from T1 = 100 to 60.

T(t) = T2 + (T1 - T2) e -kt

Solve for t, when T(t) = 60

60 = 10 + (100 - 10) e-0.01t
50/90 = e-0.01t
ln(50/90) = -0.01t
t = (ln(50/90))/-0.01
t = 58.78

Unless I've made some stupid mistake with my method (and I might have, I haven't done this sort of thing in years, and I've pieced it together from equations I've got straight off the internet), it seems that the time comes out the same.

Of course there are other things that might make a difference as well, like the specific heat capacity of water vs milk which I completely ignored for the calculation of the mixing of the liquids, as I said before milky tea may lose heat at a different rate, and as was said earlier the extra surface area once the milk has added should increase the rate of cooling. There are so many tiny variables and most of them probably make so little difference, it's very difficult to know the real life correct answer, without actually testing it.

[Gawdzilla Sama]

[Callan]

Tea, as any fule kno, is the devil's urine.
Milk only makes it slimy.

[PsychoSerenity]

I've run through the numbers again with values for the temperature of the milk below room temperature - and in this case it is faster to add the milk at the end. However, with the values I used before, and the value for the temperature of the milk at 0, it still only reached ideal temperature 3.5% faster - not necessarily enough to make up for the for the increased rate of cooling due to larger surface area if you add the milk to begin with - which was one part in five. I could try and work out how much extra surface area that gives for a cylinder - but I imagine there would be a huge difference in heat loss through the top to the air, as opposed to through the mug - so it's impossible to guess how much difference it really would make.

Maths is fun. I should do it more often.

Edit: also, what time is JimC normally about? It'd be nice to have someone check my workings.

[Warren Dew]

Unless I've made some stupid mistake with my method (and I might have, I haven't done this sort of thing in years, and I've pieced it together from equations I've got straight off the internet), it seems that the time comes out the same.

I think your calculations are correct given the assumption that the cooling rate is proportional to the difference in temperature, which would generally be true for conduction.

In actual cooling of hot water, though, heat loss through convection and evaporation are likely dominant. That means that the relationship of the speed of cooling to temperature will depend on things like the relative humidity of the air and how still or drafty the room is. In addition, things like the change in viscosity with the addition of milk might affect the rate of convective circulation within the water.

[Gawdzilla Sama]

Unless I've made some stupid mistake with my method (and I might have, I haven't done this sort of thing in years, and I've pieced it together from equations I've got straight off the internet), it seems that the time comes out the same.

I think your calculations are correct given the assumption that the cooling rate is proportional to the difference in temperature, which would generally be true for conduction.

In actual cooling of hot water, though, heat loss through convection and evaporation are likely dominant. That means that the relationship of the speed of cooling to temperature will depend on things like the relative humidity of the air and how still or drafty the room is. In addition, things like the change in viscosity with the addition of milk might affect the rate of convective circulation within the water.

Conduction would be through the cup, right? Convection into the air. Radiation would be primary for the surface of the water and secondary via the cup. So it would also depend on the thermal differential between the tea and the cup at the time the tea was introduced to that container.

[JacksSmirkingRevenge]

[MiM]

Psycho, admittedly I didn't check thorough through your calculations, but I think it is wrong to use the same k for both the tea and the mix in

T(t) = T2 + (T1 - T2) e^-kt

Clearly in normal circumstances a larger mass (tea+milk) will cool more slowly than a smaller one (only tea), so the k you are using is specific for a specific body (mass, form and material) and cannot be used as a constant in the way you do.

What people have been posting about here is only on factor in how k may be different. The probably most important factor in making up the difference in k lies in the different masses of the bodies.