Question

Using the provided data, calculate the amount of heat, in kJ, required to warm 15.3 g of solid benzene, initially at -9. °C, to gaseous benzene at 93. °C.

benzene

molar mass 78.1118 g/mol

melting point 5. °C

boiling point 80. °C

ΔHfus 9.9 kJ/mol

ΔHvap at bp 30.7 kJ/mol

Cs, solid 1.51 J/g⋅°C

Cs, liquid 1.73 J/g⋅°C

Cs, gas 1.33 J/g⋅°C

A.10.5 kJ

B.82.8 kJ

C.87.5 kJ

D.35.7 kJ

E.46.2 kJ

Please help I know what to do if it's from gas to solid but not sure about solid to gas

Answer 1

Q1 = m*Csolid*(Tf-T1)

Q2 = n*LHf

Q3 = m*Cliquid*(Tvap-Tf)

Q4 = n*LHv

Q5 = m*Cvap(T2-Tvap)

substitute all data

mol of benznee = mass/MW = 15.3/78.11 = 0.1958 mol of Benzene

Q1 = 15.3*1.51*(5--9) = 112.042 J

Q2 = n*LHf = 0.1958*9.9*1000 = 1938.42 J

Q3 = m*Cliquid*(Tvap-Tf) = 15.3*1.73*(80-5) = 1985.175

Q4 = n*LHv = 0.1958*30.7*1000 = 6011.06

Q5 = m*Cvap(T2-Tvap) = 15.3*1.33*(93-80) = 264.537

Qtotal = 112.042+1938.42+1985.175+6011.06+264.537 = 10,311.234 J = 10.31 kJ

choose A