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.

You may want to reference ( page ) Section 11.4 while completing this problem.

Part A

Calculate the heat required to convert 73.0 g of C2Cl3F3 from a liquid at 12.30 ∘C to a gas at 94.80 ∘C.
Express your answer using two significant figures.

Answer 1

heat required = m1*s1*temperature change (12.30 to 47.6 oC) + n1 * Lv + m1*s2*temperature change (47.6 oC to 94.80 oC)

where m1 = mass of  C2Cl3F3 and S1 = specific heats of C2Cl3F3(l)

s2 = specific heats of C2Cl3F3(g)

Lv = latent heat of vaporization.

Molar mass of C2Cl3F3 is 187.375 g/mol

heat of vaporization for the compound is 27.49 kJ/mol = 27.49 *1000 = 27490 J/mole

73.0 gm C2Cl3F3 = mass / molar mass = 73.0/187.375 = 0.3896 mole = n1

so put the values in above equation,

we get.

heat required = 73.0* 0.91* (47.6 -12.30) + 0.3896* 27490 + 73* 0.67 *(94.80 - 47.6 )

Heat required = 2344.9 + 10709.9 + 2308.55 = 15363.35 J = 1.5 * 10⁴ J (answer)

1.5 * 10⁴ J have two significant figures.