Answer:
q=19,767kJkJ
Explanation:
This process needs to be divided into two.
First process is actually turning it into liquid, melting and the second is bringing the temperature to the 60,6 celzius. That means that there will be two amounts of heat calculations (q1 and q2).
The things that are given, or can be read from tables are:
m=91,5g
T1=5,49°C
T2=60,6°C
ΔHfus=9,87kJ/mol, which is heat of fusion.
S=1,63J/g·°C, which is heat capacity.
First, we have to calculate the mols of the benzene we got on our hands.
It is calculated: n=91.5·(1/78.12g/mol) which equals 1,17mol.
The amount of heat needed for the first process is: q1=n·ΔHfus=1,17mol·9,87 kJ/mol= 11,548kJ
The amount of heat needed for the second process is:
q2=S·m·ΔT=1,63J/g°C·91,5g·(60,6°C-5,49°C)=8219,381J=8,219kJ
So the amount of heat needed to finish the whole process is:
q1+q2=11,548kJ+8,219kJ=19,767kJ
