Answer:
22.7
Explanation:
Initially, calculate the energy released from the sample's mass. The combustion heat represents the energy per mole of the fuel:
ΔHC=qrxnn
We can rearrange this formula to isolate qrxn, remembering to convert the sample's mass into moles:
qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J
The heat released during the reaction must match the total heat absorbed by both the water and the bomb calorimeter:
qrxn=−(qwater+qbomb)
The heat absorption by the water can be calculated using its specific heat:
qwater=mcΔT
The calorimeter's heat absorption can be derived from its heat capacity:
qbomb=CΔT
Combine both equations into the first equation, substituting the known details, with ΔT=Tfinal−20.0∘C:
−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]
Distribute each multiplication term and simplify:
−7916 J=−[(2510.4 J∘C×Tfinal)−(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)−(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)−50208 J+(420. J∘C×Tfinal)−8400 J]
Combine the similar terms and simplify:
−7916 J=−2930.4 J∘C×Tfinal+58608 J
Finally, isolate Tfinal:
−66524 J=−2930.4 J∘C×Tfinal
Tfinal=22.701∘C
Round to three significant figures gives the final result as 22.7∘C.
