A calorimeter is to be calibrated: 51.203 g of water at 55.2 degrees C is added

Question

A calorimeter is to be calibrated: 51.203 g of water at 55.2 degrees C is added to a calorimeter containing 49.783 g of water at 23.5 degrees C. After stirring and waiting for the system to equilibrate, the final temperature reached is 37.6 degrees C. Calculate the calorimeter constant.

Using the same calorimeter from the above question, calculate the amount of heat evolved when 50.00 ml of 0.400 M copper (II) sulfate solution at 23.35 degrees celcius is mixed with 50.00 ml of 0.600 M sodium hydroxide solution a same tempature. After the reaction occurs, the final tempature of the resulting mixture is measured to be 25.23 degrees celcius. The density of the final solution is 1.02 g/ml. Assume the specific heat of solution is the same as pure water, 4.184 J/g degrees celcius.

Please show the steps.

Explanation / Answer

Heat lost by hot water=mCdT=51.203*4.184*(55.2-37.6)=3766.9 J

Heat gained by cold water and calorimeter= mwater*C*dT +Ccalorimeter*dT

                                                                   =49.783*4.184*(37.6-23.5)+Ccalorimeter(37.6-23.5)

                                                                   =2934.1+14.1Ccalorimeter

Now, acc to conservation of energy.

Heat lost by hot water= Heat gained by cold water and calorimeter

3766.9=2934.1+14.1Ccalorimeter

Ccalorimeter=59.1 J/?

Now,

Volume of final mixture=50.00 +50.00=100 ml

Mass of final solution= Volume*density=100*1.02=102g

Heat evolved= msolution*Cwater*dT +CcalorimeterdT

                   =102*4.184*(25.23-23.35)+59.1*(25.23-23.35)

                   =914 J

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