Final temperature of the solution formed is 50.93ºC
As we know,
heat = mass x specific heat x change in temperature
q = mC∆T
where, m = 111 g water
C = 4.184 J/g/deg
∆T = ?
To find heat we need to first find moles Libr used in this reaction:
21.4 g Libr x 1 mole/86.84 g = 0.246 moles
Therefore,
q = 0.246 moles x (-48.8) kJ/mol = -12.00 kJ = -12000 J (since it is negative, this indicates an exothermic reaction, so temperature of solution will increase).
-12000 J = (111g)(4.184 J/g/deg)(∆T)
∆T = -25.83 degrees
Final temperature = 25.1º - (-25.83º) = 50.93ºC
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