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Imagine an idealized perfect cylindrical closed bottle which is partly filled with water. Say you have a machine which shakes the bottle horizontally for a certain time at a fixed frequency. Obviously if the bottle is almost filled or almost empty the deposited thermal energy in the water is relatively low, compared to the work performed by the shaking. For which amount of water does the temperature of the contained water reaches its maximum after shaking, starting from room temperature. One can further ask for which volume of water there is a maximum energy transfer from the mechanical motion to thermal energy.

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If you consider each part of the water in the bottle to be doing simple harmonic motion with angular frequency $\omega$ , the maximum velocity of the water is $A\omega$, where $A$ is the amplitude.

The thermal energy gained by the water would be connected to this maximum velocity, work has to be done on it each time it's 'turned around' and a proportion of that work would end up as heat in the liquid as well as creating kinetic energy $\frac{mv^2}{2}$.

When there is only a small amount of water in the bottle, $A$ is bigger, so the amount of energy going into a small mass of the water $m$ is greater than for a full bottle.

The rise in temperature is from the specific heat capacity equation

$$E= m c \Delta \theta$$

so the rise in temperature of the water is greater for a bottle without much liquid in it.

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