no of mol of ethanol gas taken(n) = 0.172 mol
mass of ethanol gas taken(m) = 0.172*46 = 7.912 g
heat released(q) = mgas*Cgas*DT1+n*DHvap+mliquid*Cliquid*DT2
mgas = 7.912 g
cgas = 1.43 j/g.c
DT1 = 300-78.5
n = 0.172 mol
DHvap = 40.5*10^3 j/mol
mliquid = 7.912 g
cliquid = 2.45 j/g.c
DT1 = 78.5-25
heat released(q) = 7.912*1.43*(300-78.5)+0.172*40.5*10^3+7.912*2.45*(78.5-25)
= 10509 joule
answer: 10509 joule
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