Answer:
22.7
Explanation:
First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:
ΔHC=qrxnn
We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:
qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J
The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:
qrxn=−(qwater+qbomb)
The heat absorbed by the water can be calculated using the specific heat of water:
qwater=mcΔT
The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:
qbomb=CΔT
Combine both equations into the first equation and substitute the known values, 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 the terms of each multiplication 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]
Add the like terms and simplify:
−7916 J=−2930.4 J∘C×Tfinal+58608 J
Finally, solve for Tfinal:
−66524 J=−2930.4 J∘C×Tfinal
Tfinal=22.701∘C
The answer should have three significant figures, so round to 22.7∘C.
