Does it make sense to compare mechanical wattages to electrical wattages? A smartphone might produce 1 watt of power. A voltage of 1 volt producing 1 ampere of current over 1 second is 1 watt of power. (Did I do that right?)
Or a kid pushing a toy block might produce 1 watt of power. Accelerating one kilogram one meter-per-second every second, moving it one meter in the direction of force, in one second — that’s also 1 watt of power.
These come from different definitions of power. In what sense does it make sense to say it is the "same" amount of power? Are they convertible or comparable?
 A: 
In what sense does it make sense to say it is the "same" amount of power? Are they convertible or comparable?

Yes, mainly due to the conservation of energy, which can be translated to a conservation of power in many situations (as power is only energy per time). Two examples:


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*In many practical situations, energy or power is eventually converted to heat. E.g., if you have a server cluster that consumes 10 kW of electrical power, it will generate 10 kW of thermal power, which you would have deduct with your cooling system.

*If it takes 1 kW of mechanical power to move a vehicle under some given conditions (speed, friction, slope), you need to feed it at least 1 kW of electrical, chemical or other power. However, since electrical and chemical power cannot be converted to mechanical power without loss, you actually need to feed it more power. The amount of loss for a given power-conversion device (motor) is actually an important characteristic, namely the energy conversion efficiency.
A: Power is simply the energy consumed (or produced, in other cases) by a component of the system. 1 watt means a dissipation of 1 joule every second, it has nothing to do with the nature of what consumes the energy. The unit of measure watt is defined exactly that way: $W = \frac{J}{s}$.
A: Power has the units  
${\rm watt} = \dfrac {{\rm joule}}{{\rm second}} =   \dfrac {{\rm newton}\times {\rm metre}}{{\rm second}} =  \dfrac {{\rm kilogramme}\times {\text{ metre second}}^{-2} \times{\rm metre}}{{\rm second}} $
${\rm watt} = {\rm volt} \times {\rm ampere} = \dfrac{{\rm joule}}{{\rm coulomb }} \times \dfrac{{\rm coulomb}}{{\rm second}} =\dfrac {{\rm joule}}{{\rm second}}$
So it is the same unit in both mechanical and electrical usage.
