Heat required(q) = n*DHfus + m*s*DT
No of mole of benzene(n) = 51.8/78 = 0.664 mol
Enthalpy of fusion of benzene(DHfus) = 9.9 kj/mol
Melting point of benzene(Tf) = 5.5 C
Mass of benzene (m) = 51.8 g
Specific heat of benzene (s) = 1.74 j/g.k
Final temperature = 40.2 c
q = 0.664*9.9*10^3+51.8*1.74*(40.2-5.5)
= 9.70 kj
answer: 9.70 kj
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