When a sample of water at 0.0°C is cooled to -36.0°C and freezes in the process,935,000 kJ of thermal energy is liberated.What is the mass of this sample of water? For water L_F = 334,000 J/kg,L_V = 2.256 × 10⁶ J/kg,and the specific heat of ice is 2050 J/kg ∙ C°.

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The correct answer is A.The mass of the water sample can be calculated using the formula:Q = mLf + mc(Tf - Ti)where Q is the thermal energy liberated, mLf is the heat of fusion of water, m is the mass of the water sample, c is the specific heat of ice, Tf is the final temperature, and Ti is the initial temperature.Substituting the given values into the formula, we get:935,000 kJ = m(334,000 J/kg) + m(2050 J/kg ∙ C°)(-36.0°C - 0.0°C)Solving for m, we get:m = 2290 kg

When a Sample of Water at 0