Asked by Sarah
A sample of KOH is dissolved into 100.0 grams of water in a coffee cup calorimeter. if the dissolution produces 6.45 kJ of heat and the temperature change is 25.4 degrees celcius to 40.0 degrees celcius, how much KOH (in grams) is added to the water? (the specific heat of the solution is 4.20 J/g degree celcius.)
Answers
Answered by
DrBob222
q = mass x specific heat x (Tfinal-Tinitial)
q = 6450 joules.
mass = only unknown
specific heat = 4.20 J/g
Tfinal = 40.0
Tinitial= 25.4
Note the correct spelling of celsius.
q = 6450 joules.
mass = only unknown
specific heat = 4.20 J/g
Tfinal = 40.0
Tinitial= 25.4
Note the correct spelling of celsius.
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