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Q.

Stearic acidCH3CH216CO2H is a fatty acid, the part of fatty that stores most of the energy. 1.0 g of stearic acid was burnt in a bomb calorimeter. The bomb had a heat capacity of 652 J/C. If the temperature of 500 g water rose from 25.0 to 39.3° C, how much heat is released when the stearic acid was burned?

[Given :CpH2O=4.18J/gC

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a

108 kJ

b

39.21 kJ

c

9.32 kJ

d

29.91 kJ

answer is A.

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Detailed Solution

For conserved energy,
Heat released = (heat capacity of bomb)ΔT+ΔT+mwCpΔT
 Here mw = mass of water in bomb calorimeter
Also,
ΔT=14.3
Thus,
H=652×14.3+500×4.18×14.3=39210.6J=39.21 kJ

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