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 10.6 g of the compound from a liquid at 30.0 °C to a gas at 60.5 °C? To be awarded credit, your solution must include a step-by-step solution with units included throughout.
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Step 1: Calculate the amount of heat required to raise the temperature of the liquid from 30.0 °C to its boiling point of 47.6 °C.
The formula to calculate the heat required to change the temperature is q = mcΔT, where q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.
ΔT = 47.6 °C - 30.0 °C = 17.6 °C
m = 10.6 g
c = 0.91 J/g-K
q1 = mcΔT = 10.6 g * 0.91 J/g-K * 17.6 K = 168.616 J
Step 2: Calculate the amount of heat required to vaporize the liquid at its boiling point.
The formula to calculate the heat of vaporization is q = nΔHvap, where n is the number of moles and ΔHvap is the heat of vaporization.
First, we need to calculate the number of moles. The molar mass of C2Cl3F3 is 2(12.01 g/mol) + 3(35.45 g/mol) + 3(18.998 g/mol) = 187.376 g/mol.
n = m/M = 10.6 g / 187.376 g/mol = 0.0566 mol
ΔHvap = 27.5 kJ/mol = 27500 J/mol
q2 = nΔHvap = 0.0566 mol * 27500 J/mol = 1556.5 J
Step 3: Calculate the amount of heat required to raise the temperature of the gas from its boiling point to 60.5 °C.
ΔT = 60.5 °C - 47.6 °C = 12.9 °C
m = 10.6 g
c = 0.67 J/g-K
q3 = mcΔT = 10.6 g * 0.67 J/g-K * 12.9 K = 90.702 J
Step 4: Add up all the heat calculated in the previous steps.
q_total = q1 + q2 + q3 = 168.616 J + 1556.5 J + 90.702 J = 1815.818 J
Therefore, the amount of heat required to convert 10.6 g of the compound from a liquid at 30.0 °C to a gas at 60.5 °C is approximately 1816 J.
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