Answer:
A)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
Explanation:
Given data :
45.00 % mole of methane
55.00 % of ethane
attached below is a detailed solution
A) calculate - Q(kw)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B ) calculate - Q ( KJ )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
since combustion takes place in a constant-volume batch reactor
