On your "separate aspect", what you're seeing is the difference in where the energy goes. I guess a little more definition may be required here. A classical setup is to have a pump drawing liquid from a stilled reservoir and pumping it somewhere. In such a case, the power input to the pump impeller shaft will go up with flow rate. Fortunately, a pump has no knowledge other than the conditions that exist at its inlet and outlet. For this reason, the same is true for an inline setup and its power also increases with flow rate.
What will vary as flow changes is how the energy put into the impeller shaft shows itself within the liquid. In the simplest sense, the "useful" work of the pump would be defined as the pressure difference on either side of the pump multiplied by the flow rate through the pump. True engineering types may complain that "work" has units of force*distance while flow rate multiplied by pressure has units of force*distance/time (power). Whatever. When you look at a centrifugal pump's efficiency curve, you'll find that they define efficiency as flow*delta-P/shaft power.
At zero flow, efficiency is zero. Likewise is true for zero delta-P. In the former, all energy goes into churning the water, ie directly into heat. You immediately see this in your example. As soon as you have some flow, the energy gets split between ineffectual churning of the water and producing flow. Efficiency climbs from zero and starts marching towards its peak. The peak efficiency will occur at some nominal flow that lies between zero flow/peak head and peak flow/nearly-zero-head. Even at peak efficiency, some of the shaft power doesn't generate any useful flow and produces heat directly. Point here is that the portion showing directly as heat is inversely related to the pump's efficiency.
The rub for a closed system is that all the energy put into the water eventually turns to heat. This is the result of pressure drops through the remainder of the system. In a truly open system you can have some energy storage if you indefinitely pull water from a low elevation and pump it to a high elevation. Note that our "open systems" really aren't open from an energy perspective.
Another thing that comes into play with submerged pumps like your example is the motor behavior. A split capacitor, single phase, four-pole motor will have a peak efficiency in the range of 65%. The efficiency curve looks like a parabola. So as you move away from the "full rated load", the efficiency will drop off dramatically. A dead head condition will tend to drop the motor output load, but will also bump the motor some along its efficiency curve. I have done absolutely zero work in this area to measure the effects, so I can't say positively how significant this is. All I would point out is that energy input to the motor does not drop as rapidly as the shaft output power drops due to the efficiency curve of the motor.
In your example above, this translates to lower total energy input when dead-headed, but 100% of that energy coverting immediately to heat. In a free-flow case, more energy is input to the fluid, but a smaller amount is directly converted to heat.
