Task about four glasses

In the comments to my post , one of the users asked an interesting question . Its essence is as follows: We have 4 glasses, with the same volume of water. 2 of them are hot and 2 are cold. Mix glasses with hot and cold water. We are waiting for 10 minutes and mix the remaining ones. Question: in which mixture will the water be hot?

Heating and cooling

Heating and cooling obey the Newton-Richman law , the solution of which has the form:

$T (t) = T_ {out} + e ^ {- kt} (T_ {0} -T_ {out})$

$T_ {out}$ - ambient temperature;
$T_ {0}$ - initial temperature;

- time;

- coefficient described below;

Let's explore this formula in detail:
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at the moment of time the temperature is equal to the initial $T_ {0}$ .
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$T (0) = T_ {out} + e ^ {- k \ cdot 0} (T_ {0} -T_ {out}) = T_ {out} + e ^ {0} (T_ {0} -T_ {out }) = T_ {out} + T_ {0} -T_ {out} = T_ {0}$
over time, it tends to be environmental (no matter whether the initial was above or below it) exponentially, at a rate proportional to the coefficient .
More details
Because $\ lim_ {t \ rightarrow \ infty} \ left e ^ {- kt} \ rightarrow 0$ then $T (t \ rightarrow \ infty) \ rightarrow T_ {out} +0 \ cdot (T_ {0} -T_ {out}) = T_ {out}$

Now a few words about the coefficient

$k = \ frac {\ alpha S} {mc}$

$\ alpha$ - heat transfer coefficient, depending on many factors, obtained experimentally.

- the area of the interface;

- weight;

- body specific heat;

If with

(contact area of water and vessel walls), specific heat capacity

and mass

everything is clear, then with the heat transfer coefficient $\ alpha$ everything is not so obvious. It shows how many joules per second will go through the square meter boundary, with a temperature difference of 1 Kelvin. Fortunately, in our case, knowing the absolute value of this coefficient is not important.

Mixing

What will be the final temperature of the water after mixing? However trite it may sound, but for our case (we mix equal amounts of water):

$T_ {H + C} = \ frac {T_ {H} + T_ {C}} {2}$

$T_ {H}$ - temperature of hot (Hot) water;
$T_ {C}$ - temperature of cold (Cold) water;
$T_ {H + C}$ - mixture temperature;

Evidence

The amount of energy released or absorbed by the system, with a change in its temperature, is related by a simple relation:

$\ Delta Q = cm \ Delta T$

Hot water, cooling to $T_ {H + C}$ will give the same amount of energy as the cold one takes, heating to the same temperature. Therefore, we can write:

$\ Delta Q_ {H} = cm (T_ {H} -T_ {H + C}) = cm (T_ {H + C} -T_ {C}) = \ Delta Q_ {C}$

$T_ {H} -T_ {H + C} = T_ {H + C} -T_ {C}$

$2T_ {H + C} = T_ {H} + T_ {C} \ Rightarrow T_ {H + C} = \ frac {T_ {H} + T_ {C}} {2}$

Glasses 2 and 4

The temperature of these glasses varies exponentially and over time $\ Delta t$ will be:

$T_ {2} (\ Delta t) = T_ {out} + e ^ {- k \ Delta t} (T_ {H} -T_ {out})$

$T_ {4} (\ Delta t) = T_ {out} + e ^ {- k \ Delta t} (T_ {C} -T_ {out})$

Next, mix and get:

$T_ {2 + 4} (\ Delta t) = \ frac {T_ {2} (\ Delta t) + T_ {4} (\ Delta t)} {2} = \ frac {T_ {out} + e ^ { -k \ Delta t} (T_ {H} -T_ {out}) + T_ {out} + e ^ {- k \ Delta t} (T_ {C} -T_ {out})} {2} =$

$= \ frac {2T_ {out} + e ^ {- k \ Delta t} (T_ {H} -T_ {out} + T_ {C} -T_ {out})} {2} = \ frac {2T_ {out } + e ^ {- k \ Delta t} (T_ {H} + T_ {C} -2T_ {out})} {2} =$

$= T_ {out} + e ^ {- k \ Delta t} \ left (\ frac {T_ {H} + T_ {C}} {2} -T_ {out} \ right)$

Glasses 1 and 3

We mix:

$T_ {1 + 3} = \ frac {T_ {H} + T_ {C}} {2}$

And through $\ Delta t$ we get:

$T_ {1 + 3} (\ Delta t) = T_ {out} + e ^ {- k \ Delta t} (T_ {1 + 3} -T_ {out}) = T_ {out} + e ^ {- k \ Delta t} \ left (\ frac {T_ {H} + T_ {C}} {2} -T_ {out} \ right)$

It would seem that there is no difference, but ...

Dirty trick or more about the k coefficient

From the point of view of heat loss, our cylindrical glass can be divided into three zones: the bottom (bottom), the upper part (top) and the side walls (sidewall). Therefore, the coefficient

can be written as an amount:

$k = k_ {b} + k_ {s} + k_ {t}$

where

$k_ {b} = \ frac {\ alpha _ {b} S_ {b}} {cm}$ , $k_ {s} = \ frac {\ alpha _ {s} S_ {s}} {cm}$ , $k_ {t} = \ frac {\ alpha _ {t} S_ {t}} {cm}$

If the glass is considered as a cylinder with a radius

and tall

, then:

$S_ {b} = S_ {t} = \ pi R ^ {2} = S$

, but

$S_ {s} = 2 \ pi Rh = 2 \ frac {S} {R} h$

We can write:

$k = \ frac {S} {cm} \ left (\ alpha _ {b} + \ frac {2h} {R} \ alpha _ {s} + \ alpha _ {t} \ right)$

So, for case 1 and 3, when mixed, the mass and height of the water column doubles:

$k_ {1 + 3} = \ frac {S} {c \ cdot 2m} \ left (\ alpha _ {b} + \ frac {2 \ cdot 2h} {R} \ alpha _ {s} + \ alpha _ { t} \ right) = \ frac {S} {cm} \ left (\ frac {\ alpha _ {b}} {2} + \ frac {2h} {R} \ alpha _ {s} + \ frac {\ alpha _ {t}} {2} \ right)$

But this has nothing to do with case 2 and 4, since they cool down (heat up) alone, and after mixing, a further change in temperature does not bother us at all.

Conclusions and another trick

If we consider

$k_ {1 + 3} = \ frac {S} {cm} \ left (\ frac {\ alpha _ {b}} {2} + \ frac {2h} {R} \ alpha _ {s} + \ frac { \ alpha _ {t}} {2} \ right)$

and

$k_ {2 + 4} = \ frac {S} {cm} \ left (\ alpha _ {b} + \ frac {2h} {R} \ alpha _ {s} + \ alpha _ {t} \ right)$

it can be concluded that $k_ {1 + 3}$ will always be less than $k_ {2 + 4}$ and therefore, if we recall that the coefficient determines the rate of temperature change, it is, in case of 1 and 3, less.

The catch is the meaning $\ frac {T_ {H} + T_ {C}} {2}$ relative to ambient temperature. If this value is higher, then the cooling of the 1 + 3 mixture will be slower than 2 + 4 and, as a result, the temperature of the first will be higher. However, if you adjust the experiment in such a way that the average temperature is lower than the ambient temperature, the 2 + 4 mixture will be “hotter”, since it heats up faster.

Interesting

If you carefully consider the coefficients, you can make another interesting conclusion.

If the cylinder is chosen in such a way that the ratio $\ frac {2h} {R}$ will be large enough (thin but long) so that $\ left (\ frac {2h} {R} \ alpha _ {s} \ gg \ alpha _ {b} + \ alpha _ {t} \ right)$ then $k_ {1 + 3} \ approx k_ {2 + 4}$ and their final temperatures will be almost the same.
On the contrary, if the ratio $\ frac {2h} {R}$ small enough (wide but short) so that $\ left (\ frac {2h} {R} \ alpha _ {s} \ ll \ alpha _ {b} + \ alpha _ {t} \ right)$ then $k_ {1 + 3} \ approx \ frac {1} {2} k_ {2 + 4}$ .
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$k_ {2 + 4} \ approx \ frac {S} {cm} \ left (\ alpha _ {b} + \ alpha _ {t} \ right)$

$k_ {1 + 3} \ approx \ frac {S} {cm} \ left (\ frac {\ alpha _ {b}} {2} + \ frac {\ alpha _ {t}} {2} \ right) = \ frac {1} {2} \ frac {S} {cm} \ left (\ alpha _ {b} + \ alpha _ {t} \ right) = \ frac {1} {2} k_ {2 + 4}$

If we substitute this value in the formula for the final temperature for 1 + 3, we get:

$T_ {1 + 3} (\ Delta t) = T_ {out} + e ^ {- \ frac {1} {2} k \ Delta t} \ left (\ frac {T_ {H} + T_ {C}} {2} -T_ {out} \ right) = T_ {out} + \ sqrt {e} \ cdot e ^ {- k \ Delta t} \ left (\ frac {T_ {H} + T_ {C}} { 2} -T_ {out} \ right)$

Recall that

$T_ {2 + 4} (\ Delta t) = T_ {out} + e ^ {- k \ Delta t} \ left (\ frac {T_ {H} + T_ {C}} {2} -T_ {out} \ right)$

If to designate:

$T_ {2 + 4} (\ Delta t) -T_ {out} = \ Delta T_ {2 + 4} = e ^ {- k \ Delta t} \ left (\ frac {T_ {H} + T_ {C} } {2} -T_ {out} \ right)$

and

$T_ {1 + 3} (\ Delta t) -T_ {out} = \ Delta T_ {1 + 3} = \ sqrt {e} \ cdot e ^ {- k \ Delta t} \ left (\ frac {T_ { H} + T_ {C}} {2} -T_ {out} \ right) = \ sqrt {e} \ cdot \ Delta T_ {2 + 4}$

Thus, for an arbitrary cylindrical cup and $\ Delta t$ , $\ left | \ Delta T_ {1 + 3} \ right |$ will lie between $\ left | \ Delta T_ {2 + 4} \ right |$ and $\ sqrt {e} \ cdot \ left | \ Delta T_ {2 + 4} \ right |$ :

$\ left | \ Delta T_ {2 + 4} \ right | <\ left | \ Delta T_ {1 + 3} \ right | <\ sqrt {e} \ cdot \ left | \ Delta T_ {2 + 4} \ right |$

And in fact..?

In fact, this article is a very clear example of the fact that it is impossible to approach practice without a theory. And this paragraph contains conclusions more importantly than the above. Not knowing the fact that the result depends, among other things, on the ambient temperature, will lead to misinterpretation of the latter. Just imagine, two friends set up an experiment, cold water 5 degrees, hot 60. But one of them is at home 25, and the other 40. The results will be inconsistent. And if they also have different glasses or actions are not synchronous ... But much worse, it is false to say about the only correct outcome of the experiment, if the results are the same (due to the fact that hot water is more often "hotter" than cold "cooler" relative to room temperature ). Also, you should always at least approximately estimate the output values. If, for example, the difference in temperatures is quantitatively 0.5 degrees, then it is foolish to measure it with a room thermometer, with an error of 2 degrees. It is worth mentioning that the Newton-Richman law is valid only if the heat in the water spreads much easier than through the walls of the glass. In addition, the thermal conductivity, as well as the specific heat capacity, may depend on temperature. Well, our conclusions for the cylindrical glass, and other geometry will make its own adjustments.

Source: https://habr.com/ru/post/365325/

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