91Ó°ÊÓ

Before we can calculate the change in gravitational potential energy, we need to know the gravitational constant, which is approximately \(9.81 \; \mathrm{m/s^2}\). The change in gravitational potential energy can be calculated using the formula: $$\Delta E_p = m \cdot g \cdot h$$ Where \(\Delta E_p\) is the change in gravitational potential energy, m is the mass of the falling water, g is the gravitational constant, and h is the height of the fall. The mass m of the falling water is \(1.00 \; \mathrm{kg}\), and the height h is \(0.100 \; \mathrm{km} = 100 \; \mathrm{m}\). Plug in the values and calculate the change in potential energy: $$\Delta E_p = 1.00 \; \mathrm{kg} \cdot 9.81 \; \mathrm{m/s^2} \cdot 100 \; \mathrm{m}$$ $$\Delta E_p = 981 \; \mathrm{J}$$

The specific heat of water is approximately \(4.18 \; \mathrm{J/(g \cdot ^\circ C)}\). The temperature change of the mixed water is \(0.1 ^\circ \mathrm{C}\). The mass of the water initially in the vessel is unknown; let's call it M. The heat energy gained by the mixed water can be calculated as follows: $$Q = (m_1 + M) \cdot C_p \cdot \Delta T$$ Where Q is the heat energy gained, \(m_1\) is the mass of the falling water, M is the mass of the water in the vessel, \(C_p\) is the specific heat of water, and \(\Delta T\) is the temperature change of the mixed water.

Since the energy gained from the fall is converted into heat energy upon mixing with the water, we can write the following equation: $$\Delta E_p = Q$$ $$981 \; \mathrm{J} = (1.00 \; \mathrm{kg} + M) \cdot 4.18 \; \mathrm{J/(g \cdot ^\circ C)} \cdot 0.1 ^\circ \mathrm{C}$$ $$981 \; \mathrm{J} = 0.418 \; \mathrm{J/^{\circ}C} \cdot (1.00 \; \mathrm{kg} + M)$$ Now, solve for M: $$M = \frac{981 \; \mathrm{J} - 0.418 \; \mathrm{J/^{\circ}C} \cdot 1.00 \; \mathrm{kg}}{0.418 \; \mathrm{J/^{\circ}C}}$$ $$M \approx 2340.19 \; \mathrm{g} = 2.34 \; \mathrm{kg}$$ So, the mass of the water initially present in the vessel is approximately \(2.34 \; \mathrm{kg}\).