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

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge