Answer : The energy removed must be, -34.67 kJ
Solution :
The process involved in this problem are :
[tex](1):C_6H_6(l)(425.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(2):C_6H_6(l)(353.0K)\rightarrow C_6H_6(s)(353.0K)\\\\(3):C_6H_6(s)(353.0K)\rightarrow C_6H_6(s)(335.0K)[/tex]
The expression used will be:
[tex]\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = heat available for the reaction = [tex]4.50\times 10^3kJ=4.50\times 10^6J[/tex]
m = mass of benzene = 125 g
[tex]c_{p,s}[/tex] = specific heat of solid benzene = [tex]1.51J/g.K[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid benzene = [tex]1.73J/g.K[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]9.8kJ/mole=9800J/mole=\frac{9800J/mole}{78g/mole}J/g=125.64J/g[/tex]
Molar mass of benzene = 78 g/mole
Now put all the given values in the above expression, we get:
[tex]\Delta H=[125g\times 1.73J/g.K\times (353-425)K]+125g\times -125.64J/g+[125g\times 1.51J/g.K\times (335-353)K][/tex]
[tex]\Delta H=-34672.5J=-34.67kJ[/tex]
Therefore, the energy removed must be, -34.67 kJ
