Question

The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 ∘C. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/g⋅K and 0.67 J/g⋅K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.

Calculate the heat required to convert 51.0 g of C2Cl3F3 from a liquid at 10.90∘C to a gas at 88.70∘C Express your answer using two significant figures.
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Answer 1

The heat required to convert 51.0 g of C2Cl3F3 from a liquid at 10.90°C to a gas at 88.70°C is 1.05 × 10^4 J.

Step 1: Calculate the heat required to raise the temperature of liquid C2Cl3F3 from 10.90°C to 47.6°C

q_liquid = m * c_liquid * ΔT = 51.0 g * 0.91 J/g·K * (47.6°C - 10.90°C) = 1657.56 J

Step 2: Calculate the heat required to vaporize 51.0 g of C2Cl3F3

q_vaporization = n * ΔHvap = (51.0 g / 183.33 g/mol) * 27.49 kJ/mol = 7.53 kJ

(Note: 183.33 g/mol is the molar mass of C2Cl3F3)

Step 3: Calculate the heat required to raise the temperature of gaseous C2Cl3F3 from 47.6°C to 88.70°C

q_gas = m * c_gas * ΔT = 51.0 g * 0.67 J/g·K * (88.70°C - 47.6°C) = 1284.66 J

Step 4: Calculate the total heat required

q_total = q_liquid + q_vaporization + q_gas = 1657.56 J + 7530 J + 1284.66 J = 10472.22 J

Step 5: Round the answer to two significant figures

q_total = 1.05 × 10^4 J

Therefore, the heat required to convert 51.0 g of C2Cl3F3 from a liquid at 10.90°C to a gas at 88.70°C is 1.05 × 10^4 J.