The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 degree celsius The specific heats of Cl3F3(l) and …
ID: 953713 • Letter: T
Question
The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 degree celsius The specific heats of Cl3F3(l) and C2Cl3F3(g) are 0.96 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 50.0 g of fluorocarbon at 10.0 C degree to fluorocarbon at 85.00 C degree
The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 degree celsius The specific heats of Cl3F3(l) and C2Cl3F3(g) are 0.96 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 50.0 g of fluorocarbon at 10.0 C degree to fluorocarbon at 85.00 C degree
Explanation / Answer
We need to calculate three different heats and add them together. First, we need to calculate how much heat was required to warm the liquid from 10.00 C to its boiling point at 47.6 C:
q1 = m c DT = 50.0 g X 0.96 J/gC X 37.6 C = 1804J
Then calculate the heat required to vaporize that mass of the compound. Since the heat of vaporization is expressed in kJ/mol, you can either divide this by the molar mass of the compound to convert the heat of vaporization into kJ/g OR, you can calculate the number of moles of the compound in 10 grams. The molar mass of the compound is 187.3, so 50 grams = 0.267 moles.
So,
q2 = moles X Hvap = 0..267 X 27.49 kJ/mol = 7.336 kJ = 7336 J
Finally, calculate the heat required to warm the gas from 47.6 C to 85 C:
q3 = m c DT = 50.0 X 0.67 X 37.4 = 1253 J
Adding those three together i.e., q1+q2+q3 =1804+7336+1253 = 10,393 J = 10.39 kJ