How much energy is required to increase the temperature of 36.0 g of H2O from - 20 °C to 50 °C, keeping in mind the following: Vice ≈ 2.00 J/g*C Liquid water ≈ 4.00 J/g*C ΔHfusion ≈ 6000 J/mol

2.06 x 10⁴ J

Explanation:

The process takes place in three steps. First, the ice is heated from - 20 °C to 0 °C. Then the ice undergoes a phase change to water. Finally, the water is heated from 0 °C to 50 °C.

The heat energy required for the first step is as follows:

Q = mcΔT = (36.0 g) (2.00 Jg⁻¹°C⁻¹) (0 °C - (-20 °C)) = 1440 J

The heat energy required for the phase change (where L is the heat of fusion) is then calculated. Grams are converted to moles using the molar weight of water (18.02 g/mol)

Q = ML = (36.0 g) (mol/18.02g) (6000 J/mol) = 11987 J

Finally, the heat energy required to raise the temperature of the water to 50°C is calculated:

Q = mcΔT = (36.0 g) (4.00 Jg⁻¹°C⁻¹) (50 °C - 0 °C) = 7200 J

Adding all of the heat energy values together gives:

(1440 + 11987 + 7200) J = 20627 J

The final answer is 2.06 x 10⁴ J