Chapter 14 - Fluids - Problems - Page 408: 21

0.635 J

Let $h_1$ and $h_2$ be the initial heights of the liquid in the two vessels. If $h$ is the height of the liquid in the vessels after they are connected, we can write $h={h_1+h_2\over 2}={1.560+0.854\over 2}=0.353\,\mathrm m$. When the connection between the vessels is made, the volume of the liquid in vessel A of height $h_1-h$ will appear in vessel B. This volume is $V=A(h_1-h)$. The mass of the liquid displaced is $m=\rho V=\rho A (h_1-h)$. The height through which the liquid drops in vessel A is $(h_1-h)$. The work done by the gravitational force in this process is then \begin{align*} W&=mg(h_1-h)\\ &=\rho Ag(h_1-h)^2\\ &=1.30\times 10^3\times 4.00\times 10^{-4}\times 9.80 \times (0.353)^2\\ &=0.635\,\mathrm J \end{align*}