A) 21 kJ
B) 16 kJ
C) 13 kJ
D) 62 kJ
Correct Answer: D
Solution :
\[\Delta H=n\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{{{C}_{p.m}}dT}=3\times \int\limits_{300}^{1000}{(23+0.01T)dT}\] \[=3[23(1000-300)+\frac{0.01}{2}({{1000}^{2}}-{{300}^{2}})]\] \[=61950\,J\approx 62kJ\] Correct option : (d)You need to login to perform this action.
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