Calculate the energy that is required to heat 125 g ice at -10.0°C to a temperature of 130.0°C.The heat of fusion = 333 J/g,the heat of vaporization = 2256 J/g,and the specific heat capacities of ice = 2.06 J /g∙K,liquid water = 4.184 J /g∙K,and steam = 2.01 J /g∙K.

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The total energy required is the sum of the energy needed to heat the ice from -10°C to 0°C, melt the ice, heat the water from 0°C to 100°C, vaporize the water, and heat the steam from 100°C to 130°C. Calculating each step:  1. Heating ice: q1 = m * c_ice * ΔT = 125 g * 2.06 J/g∙K * 10 K = 2575 J  2. Melting ice: q2 = m * heat of fusion = 125 g * 333 J/g = 41625 J  3. Heating water: q3 = m * c_water * ΔT = 125 g * 4.184 J/g∙K * 100 K = 52300 J  4. Vaporizing water: q4 = m * heat of vaporization = 125 g * 2256 J/g = 282000 J  5. Heating steam: q5 = m * c_steam * ΔT = 125 g * 2.01 J/g∙K * 30 K = 7537.5 J  Total energy = q1 + q2 + q3 + q4 + q5 = 387037.5 J ≈ 3.87 × 10^5 J.

Calculate the Energy That Is Required to Heat 125 G