Asked by Chris
The heat of solution of ammonium chloride is 15.2 kJ/mol. If a 6.187 g sample of NH4Cl is added to 65.0 mL of water in a calorimeter at 26.5°C, what is the minimum temperature reached by the solution? [specific heat of water = 4.18 J/g°C; heat capacity of the calorimeter = 365. J/°C]
Answers
Answered by
bobpursley
The sum of heats gained is zero.
heatgainedwater+heataddedtoNH3Cl=0
65g*cwater*(Tf-26.5)+6.187/molmass*15.2kj/mol=0
now, change c water to Kj/g to make units work out (kJ)
cwater=.004kJ/gC
and that ought to let you work out what is tf.
heatgainedwater+heataddedtoNH3Cl=0
65g*cwater*(Tf-26.5)+6.187/molmass*15.2kj/mol=0
now, change c water to Kj/g to make units work out (kJ)
cwater=.004kJ/gC
and that ought to let you work out what is tf.
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