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If I am at a cold place, can I warm a bottle of water just by continuously shaking it?

If yes, how long would it approximately take, i.e., is it physically possible or just theoretically?

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    $\begingroup$ The last post addresses your question: reddit.com/r/askscience/comments/2zekje/… $\endgroup$ – user198207 Jun 22 '18 at 11:49
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    $\begingroup$ The answer is yes. Viscous deformation of the water causes its internal energy and temperature to increase. The answer to your second question is that it will start to warm ever so slightly as soon as you start shaking it. $\endgroup$ – Chet Miller Jun 22 '18 at 12:08
  • $\begingroup$ Are you shaking it with a robot arm? Just touching it will "heat it" since you are hotter than the water (assuming the bottle is in equilibrium with the cold place). How hot do you want the water to be? $\endgroup$ – ggcg Jun 22 '18 at 12:08
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    $\begingroup$ In order for viscous deformation to contribute to the thermal balance on the fluid, the fluid has to actually be deforming. This won't happen if the fluid is rotating as a rigid body. $\endgroup$ – Chet Miller Jun 22 '18 at 15:01
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    $\begingroup$ what-if.xkcd.com/71 $\endgroup$ – John Jun 22 '18 at 18:31
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Let us assume you shake from each side at a rate $f$ of say four times per second attaining $\Delta v/2=5\ \mathrm{mph}$ $\simeq$ 8 kph before reversing directions, and that some fraction of the energy $\beta$ you give each switch is converted into heat which is absorbed by the water, and that none of it is radiated out.

The value of $\beta$ will change based on how full the bottle is. If the bottle is 100% full, then there will be no internal motions for viscosity to convert the energy into heat, and $\beta\simeq0$.

To understand this situation (the $\beta\simeq0$ full-bottle case), one could "heat" the water by putting the bottle into a simple harmonic oscillator (with a spring), and it would go back and forth without heating the water and without any expenditure of energy. By conservation of energy, no heat can be added to the water.

Continuing, the desired change in energy is

$$\Delta E=M_\text{water}C_V\Delta T,$$

where $M_\text{water}$ is the mass of the water, $C_V$ is the heat capacity of water, and $T$ is the temperature, with $\Delta$ denoting changes.

Meanwhile, the total heat supplied will be

$$\Delta E=N \beta {1 \over 2}M_\text{water}{\Delta v}^2,$$

where $N$ is the number of single-direction shakes.

Solving for $N$, we get

$$N={2C_V\Delta T \over {\beta \Delta v}^2}.$$

The total time it takes will be

$$T={N\over f}={2C_V\Delta T \over \beta f{\Delta v}^2}.$$

If warm is $100\ \mathrm{^\circ F}\simeq38\ \mathrm{^\circ C}$ and cold is $32\ \mathrm{^\circ F}=0\mathrm{^\circ C} $, $\Delta T\simeq68\ \mathrm{^\circ F}\simeq38\ \mathrm{^\circ C}$. 5 mph $\simeq$ 2.2 m/s. So, if $\beta=0.5$,

$$T={2\times 4184\ \mathrm{\frac{J}{kg\ K}} 38\ \mathrm K \over 0.5 \times 4\ \mathrm s^{-1} \times \left(2\times2.2\ \mathrm{m\over s}\right)^2}=8212\rm \space seconds\simeq 2 \space hours \space 16 \space minutes.$$

This doesn't account for radiative losses during that time, although that won't become too important unless you want to make tea. Actually, if you wanted boiling hot water for that purpose, then you could have made such tea in a few minutes of shaking (without changing the temperature much) because the motion will drastically increase the effective diffusion rate of liquid through the tea leaves vs. stagnant water.

I'd like to thank Loong for syntax corrections and also Ujjwal Barman for asking a question that made me want to include a $\beta$ factor, as the original answer had none and I'd hate to have someone shake a full bottle for hours only to find disappointment!

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  • $\begingroup$ if i am correct, you've considered that at those moments where i reverse the direction of motion of the bottle the kinetic energy of the water molecules inside all get converted to heat. but for that to be true they'll have to come to rest. I dont think thats what will happen. imagining a simple situation, if a single ball collides with a wall, only time its KE is converted wholly into heat is when it sticks to the wall, not when it bounces back. if you see what i am getting at.. $\endgroup$ – Ujjwal Barman Jun 22 '18 at 14:40
  • $\begingroup$ You are largely correct. I've added a factor of beta to account for the elastic nature of accelerating the water and gave you credit for the idea. Sorry for the delays as there was a power outage in the middle of me editing and I'm still learning how this site works. $\endgroup$ – Jason Arthur Taylor Jun 22 '18 at 17:25
  • $\begingroup$ I think the important question is what proportion of shaking energy goes to the internal energy of the water? First, I'd argue the water is nearly frictionless, and mostly elastic collisions, meaning very little of the energy actually converts to heat. Second, most of the energy goes into changing direction of the vessel--not actually changing the internal energy of the water, when you shake it +x, you then shake it -x. $\endgroup$ – James Jun 22 '18 at 20:14
  • $\begingroup$ James, you're arguing for a low $\beta$ without saying so. For simplicity, I used 0.5, but I agree it could be lower. What value do you think is better? $\endgroup$ – Jason Arthur Taylor Jun 22 '18 at 20:32
  • $\begingroup$ Turbulence is an efficient energy sink. Given how well a water bottle (doesn't) bounce, 0.5 seems very conservative. $\endgroup$ – BowlOfRed Jun 22 '18 at 22:50
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A bottle closed at one end with a bung at the other end has some water in it.

The bung has a thermometer passing through it to measure the temperature of the water.

The bottle, initially in a vertical position, is repeatedly inverted.

With no heat losses you can equate the loss in gravitational potential energy to the gain in heat energy.

$$mg\Delta h N= mC\Delta \theta$$

where $m$ is the mass of a small amount of water in the bottle of length $\Delta h$ which is inverted $N$ times, $g$ is the gravitational field strength, $C$ the specific heat capacity of water and $\Delta \theta$ the temperature rise.

$$\Rightarrow \Delta \theta = \dfrac{9.8 \times 0.3}{4200} N= 0.0007 \, N$$

for a bottle of length $0.3\,\rm m $

So about $1400$ inversions should raise the temperature by $1\, ^\circ \rm C$.

Heat loss and the thermal capacity of the bottle will mean that many more inversions will be necessary.
Also one must insulate the bottle so that the process of inversion using hands does not heat the water.

So let it be $3\times 1400 = 4200 $ inversions.

One inversion every second gives a time of about $70$ minutes to gain $1\, ^\circ \rm C$ so perhaps it should be a $0.2\, ^\circ \rm C$ temperature rise and a more reasonable time of approximately $15$ minutes.


Update after doing two experiments

enter image description here

The first experiment I tried was to use lead shot in a tube of length $0.77 \,\rm m$ with some lead shot in it.
I measured the initial temperature of the lead shot $(24.3 \rm ^\circ C)$ and the final temperature after $100$ inversions $(25.2 \rm ^\circ C)$.
Assuming no heat losses, 100% conversion of gravitational potential energy into heat etc this gives an unexpected specific heat capacity of lead of about $800\, \rm J\,kg^{-1}\,K^{-1}$.
Unexpected because the documented value for the specific heat capacity of lead is $160\, \rm J\,kg^{-1}\,K^{-1}$ but also because a report found by @JasonArthurTaylor using the same method found the specific heat capacity of lead to be approximately $800\, \rm J\,kg^{-1}\,K^{-1}$.
This seems to indicate that although the method does show an increase in temperature the temperature rise is well below the value predicted using a simple theory.

Although the result of the first experiment was unexpected I did not have time to repeat it because I wanted to address the question posed by the OP regarding a bottle of water.

To a $2$ litre plastic bottle was added about $200\,\rm cm^3$ of water and a thermometer with $0.1^\circ\rm C$ graduations was used to measure the temperature of the water at one minute intervals.
The bottle was shaken at a rate of approximately $170$ shakes per minute over a distance of about $15\,\rm cm$
All I wanted to find out is the order of magnitude of the temperature rise.

The water temperature before the shaking started was $23.8 ^\circ \rm C$ and the water temperature at one minute intervals up to $5$ minutes was $24.0 ^\circ \rm C$, $24.4 ^\circ \rm C$, $24.6 ^\circ \rm C$, $24.9^\circ \rm C$ and $25.1 ^\circ \rm C$.

So this is good evidence that shaking a water bottle does increase the temperature of the water inside the bottle.

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  • $\begingroup$ "With no heat losses you can equate the loss in gravitational potential energy to the gain in heat energy." I question this. If generally true, would not this youtube.com/watch?v=1feybxNChU0 slow down in 1 revolution if tilted at 90$^\circ$? $\endgroup$ – Jason Arthur Taylor Jun 25 '18 at 16:15
  • $\begingroup$ @JasonArthurTaylor I have suggested that the is “some (a little) water” in the bottle so the water falls the length of the bottle. In the video the jars are (almost) full of water. What I have described is the basis of a Physics demonstration using lead shot. physics.usyd.edu.au/super/therm/tpteacher/demos/shottube.html The problem with having a container nearly full of water is making the water move. $\endgroup$ – Farcher Jun 25 '18 at 17:54
  • $\begingroup$ But your word $some$ does not actually exclude $full$, as you imply. Lead is unusual, and was selected for a reason. It depends upon how rapidly the collisions occur in relation to the break up self acceleration limit. If it is too low, there won't be the energy conversion. $\endgroup$ – Jason Arthur Taylor Jun 25 '18 at 23:32
  • $\begingroup$ @JasonArthurTaylor If you read further on I use the phrase “small amount of water”. $\endgroup$ – Farcher Jun 25 '18 at 23:36
  • $\begingroup$ Was not your sole rationale for your use of this word regarding the approximation of ignoring the height of the water as compared to $\Delta h$, the height of the bottle? I'd guess yes because your use of "small" would have probably accompanied "some" instead of $\Delta h$. Now, you are implying that you already knew of the lack of turbulence issue that I had already pointed out in my answer, but the location of "small" in your answer IMO does not support this position. $\endgroup$ – Jason Arthur Taylor Jun 26 '18 at 1:41

protected by Qmechanic Jun 22 '18 at 22:02

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