The problem involves energy transfer and phase change from a solid (ice) to a liquid (water), so we use these concepts to solve it.
Firstly, given that the energy transferred to ice is 9.10 * 10^5 J and the latent heat of fusion is 3.33 * 10^5 J/kg, we can calculate the mass of ice that melts by dividing these two values.
Consequently, we find that 1.8 kg of ice melts.
We then check if all ice has been melted. Since the mass of ice that has been melted is equal to the total mass of ice, we conclude that all the ice has melted.
Having ascertained this, we need to find the final temperature of the system. The excess energy, which results from the transfer process after all the ice has melted, is used to increase the temperature of the liquid water.
The excess energy, which is the energy used to melt the ice subtracted from the total energy transferred, is (1.8 - 1.8) * 3.33 * 10^5 J/kg = 0 J.
The final temperature, calculated by adding the initial temperature (0°C) to the ratio between the excess energy and the product of the mass and specific heat of water (4186 J/(kg°C)), is 0 + 0 / (1.8 * 4186) = 0°C.
But, since 0 J is the excess energy after all ice has been melted, there is a fact that there is still some energy left that will be used to heat the water, so the calculation should be slightly adjusted. Instead, the excess energy should equal the total energy transferred minus the energy needed to melt all the ice, equals 9.10 * 10^5 J - 1.8 * 3.33 * 10^5 J/kg. The final temperature is then calculated as before and found to be approximately 41.222°C.
So, all of the 1.80 kg of ice melts, and the eventual temperature of the water is about 41.222°C.