1. 0.25-mol ice at -5 °C is mixed with n-mol hot water initially at 45 °C...

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1. 0.25-mol ice at -5 °C is mixed with n-mol hot water initially at 45 °C in an isobaric adiabatic calorimeter at 1 atm. The

Answer 1

Answer #1

heat gained by ice(q) = n*c1*DT1 + n*DHfus + n*c2*DT2

n = no of mol of ice = 0.25 mol

c1 = molar heatcapacity of ice = 38.09 j/mol.k

DT1 = 5 c

DHfus = 6.008 kj/mol

n = no of mol of ice converted into water = 0.25 mo l

c2 = molar heatcapacity of ice = 75.291 j/mol.k

DT2 = 10 c

q = 0.25*38.09*5+0.25*6.008*10^3+0.25*75.291*10

= 1.74 kj

DHrxn = +q

= 1.74 kj

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Answer 2

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