Asked by JD
The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 °C. The specific heat of this compound in the liquid state is 0.91 J/g-K and in the gas state is 0.67 J/g-K. The heat of vaporization is 27.5 kJ/mol. What is the amount of heat required to convert 5.6 g of the compound from a liquid at 30.0 °C to a gas at 60.5 °C?
Answers
Answered by
DrBob222
q within the liquid state from T of 30 C to T of 47.6 is
q = mass x specific heat liquid x (Tf-Ti) or
q1 = 5.6 g x 0.91 J/g.K x (47.6 - 30) = ? J
q2 to convert liquid to gas @ the boiling point is
q2 = delta H x mol = 27.5 kJ/mol x 5.6g/molar mass = ? kJ
q3 within the gas phase is
q3 = mass x specific heat x (Tfinal-Tinitial) =
q3 = 5.6 g x 0.67 J/g.K x (60.5 - 47.6) = ? in J
Total = qtotal = q1 + q2 + q3 = ?
q = mass x specific heat liquid x (Tf-Ti) or
q1 = 5.6 g x 0.91 J/g.K x (47.6 - 30) = ? J
q2 to convert liquid to gas @ the boiling point is
q2 = delta H x mol = 27.5 kJ/mol x 5.6g/molar mass = ? kJ
q3 within the gas phase is
q3 = mass x specific heat x (Tfinal-Tinitial) =
q3 = 5.6 g x 0.67 J/g.K x (60.5 - 47.6) = ? in J
Total = qtotal = q1 + q2 + q3 = ?
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