How much pressure do I need to apply to a bottle of water to heat it? Let's say I have a normal 0.5 l plastic water bottle.
Let's say the room and the bottle/water are at 20 °C / 1 atm.
How much pressure do I need to apply to the bottle so that the water raises in temperature, let's say, by 1 °C? 
What about 10 °C?
Could I be able to do it with my bare hands?
 A: It would require extremely high pressures, pressures that would readily rupture the bottle, before you would be able to raise the temperature of water even a tiny amount. That is because water is relatively incompressible. 
If you grab the bottle by your bare hands, you will raise the temperature of the water more by heat transfer from your hands to the water in the bottle, than by any conceivable pressure you could apply. 
Hope this helps.
A: This analysis addresses the problem of the temperature change resulting from compressing liquid water without any air present.  If the work is done adiabatically and reversibly, then, from the first law, per mole $$dU=-PdV$$ and, with the differential change in internal energy given by the fundamental equation:  $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$Therefore, $$C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV=-PdV$$From this, if follows that $$C_vdT=-T\left(\frac{\partial P}{\partial T}\right)_VdV$$In addition, the following mathematical relationship applies:  $$\left(\frac{\partial P}{\partial T}\right)_V=-\frac{\left(\frac{\partial V}{\partial T}\right)_P}{\left(\frac{\partial V}{\partial P}\right)_T}$$The equation of state for a liquid is typically expressed as:  $$\frac{dV}{V}=\alpha dT-\beta dP$$where $\alpha$ is the volumetric coefficient of thermal expansion and $\beta$ is the bulk compressibility:  $$\alpha=\left(\frac{\partial \ln{V}}{\partial T}\right)_P$$ $$\beta=-\left(\frac{\partial \ln{V}}{\partial P}\right)_T$$
If we combine the previous equations, we obtain:$$C_vdT=-TV\frac{\alpha}{\beta}\left(\alpha dT-\beta dP\right)$$ or $$\left(C_v+TV\frac{\alpha^2}{\beta}\right)dT=TV\alpha dP$$
So, as a final result, $$\frac{dT}{dP}=\frac{TV\alpha}{C_v+TV\frac{\alpha^2}{\beta}}$$or $$\Delta P=\left(\frac{\alpha}{\beta}+\frac{C_v}{\alpha TV}\right)\Delta T$$I leave it up to you to look up the molar heat capacity, molar volume, coefficient of volume expansion, and bulk compressibility of water at room temperature 293 K.
A: From a perspective of thermodynamics:
Suppose there's 1mL gas in the bottle and there's no heat transfer between water bottle and surroundings. Then the bottled water could be considered as a 500 mL liquid phase/ 1 mL gas phase in equilibrium in a closed system. Also please notice that in this system, the gas phase is not pure water vapor, but a mixture of air and a tiny amount of water vapor, therefore it can't be described by Clausius- Clapeyron equation. 
Isochoric specific heat capacity of water at 25 degree C is 4.13kJ/kg*K, so the energy it takes to rise a 500mL bottled water 1 K is about 2065 J.
Now if we compare the initial state and final state of the system:
Assume water is incompressible(I think it's reasonable since water compressibility is ~5x 10^-10 Pa^-1), then liquid phase volume change =0 mL;
Majority of gas phase disappears (most water vapor undergoes a phase transition and becomes liquid, while most other gas either dissolves in water or is compressed to a negligible volume), then gas phase volume change is approximately 1mL.
Hence total volume change "delta V" is the summation of volume changes of gas phase and liquid phase, which is approximately 1 mL. 
The work done to the system then roughly equals to 
P x delta V = 2065 J.   
(Condensation heat of water vapor is neglected here because it's an extremely small number compared with 2065J)
P needed is approximately 2065kPa, which is about 20 atmosphere pressure. At least I can't do it by my bare hands.  :)
