Does shaking a water bottle increase the temperature of the water inside? 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?
 A: 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 

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.
A: 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!
