Problem Analysis
The problem states the ice block ($V = 100 \text{ cm}^3$) is placed in Vessel A and becomes half-submerged. Since ice naturally floats with about 90% submersion, being only 50% submerged implies the water level is shallow, and the ice block is resting on the bottom of the container.
Step 1: Calculate Initial Displacement
The ice block has a total volume $V_{ice} = 100 \text{ cm}^3$. Since it is half-submerged, the volume of water currently displaced by the ice is:
$$V_{disp} = \frac{1}{2} \times V_{ice} = \frac{1}{2} \times 100 \text{ cm}^3 = 50 \text{ cm}^3$$Step 2: Calculate Volume of Water from Melted Ice
Mass is conserved during melting. We find the mass of the ice ($m_{ice}$) using its density ($\rho_{ice} = 900 \text{ kg/m}^3 = 0.9 \text{ g/cm}^3$):
$$m_{ice} = V_{ice} \times \rho_{ice} = 100 \text{ cm}^3 \times 0.9 \text{ g/cm}^3 = 90 \text{ g}$$The volume of water produced ($V_{melt}$) when this mass becomes water ($\rho_{water} = 1 \text{ g/cm}^3$) is:
$$V_{melt} = \frac{m_{ice}}{\rho_{water}} = \frac{90 \text{ g}}{1 \text{ g/cm}^3} = 90 \text{ cm}^3$$Step 3: Determine Flow
The ice effectively occupied a volume of $50 \text{ cm}^3$ in the water. After melting, it adds $90 \text{ cm}^3$ of liquid water to the system.
The excess volume added to the system is:
$$\Delta V = V_{melt} – V_{disp} = 90 – 50 = 40 \text{ cm}^3$$Since the vessels are identical and connected, the water levels will equalize. The excess volume distributes equally between the two vessels. Therefore, half of the excess flows from Vessel A to Vessel B.
$$V_{flow} = \frac{\Delta V}{2} = \frac{40 \text{ cm}^3}{2} = 20 \text{ cm}^3$$Mass of water flowing = $20 \text{ g}$.