A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb calorimeter is 420. J∘C and the heat of combustion at constant volume of the sample is - 3374kJmol, calculate the final temperature of the reaction in Celsius. The specific heat capacity of water is 4.184 Jg ∘C.

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.