How much energy must be removed from a 94.4 g sample of benzene at 322.0K to solidify the sample and lower the temp to 205.0 K?

Answer 1

Known info:

mass Benzene = 94.4

ΔHfus = 9.8 kj/mol

Specific Heat liquid = Cliq = 1.73 J/g

Specific Heat Solid = Csolid = 1.51 J/g

Temp Melting = 279.0 K

3 Steps

A) q = (mass)(Cliq)(ΔT)

for change in temp use (322- 279), you're going from the highest temp of 322 K to the point the benzene would melt at 279 K

q = (94.4)(1.73)(43)

q = 7022.416 convert to kj = 7.022 kj

B) use q = (n)( ΔHfus)

molar mass = 78.11

convert to moles => 94.4/78.11 = 1.21

q=(1.21)(9.8)

q=11.8 kj

C) q = (mass)(Csolid)(ΔT)

for change in temp here you're going from the melting point to the low temp given. use (279-205)

q = (94.4)(1.51)(74)

q = 10548.256 convert to kj => 10.548 kj

now add all 3 steps together.

A + B + C = final answer

(7.022 kj) + (11.8 kj) + (10.548 kj) = 29.4 kj