A static, uniform liquid is on one side of a double container as shown in \(abla\) Figure 5.19a. If the valve is open, the level will fall, because the liquid has (gravitational) potential energy. This may be computed by assuming all the mass of the liquid to be concentrated at its center of mass, which is at a height \(h / 2\). (More on the center of mass in Chapter 6.5.) When the valve is open, the liquid flows into the container on the right, and when static equilibrium is reached, each container has liquid to a height of \(h / 2\), with centers of mass at \(h / 4\). This being the case, the potential energy of the liquid before opening the valve was \(U_{\mathrm{o}}=(m g) h / 2\), and afterward, with half the total mass in each container (Figure 5.19b), \(U=(m / 2) g(h / 4)+(m / 2) g(h / 4)=\) \(2(m / 2) g(h / 4)=(m g) h / 4\). Whoa. Was half of the energy lost?

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Closed valve Opened valve 2 (0) FIGURE 5.19 Is energy lost? See Conceptual Example 5.10 for description. (b)