In an experiment to determine the enthalpy of neutralization using the calorimet
ID: 517671 • Letter: I
Question
In an experiment to determine the enthalpy of neutralization using the calorimeter the lab was performed into two parts. In PART (A), they put 20 mL of water in the calorimeter. After it reaches thermal equilibrium with the calorimeter, they measure its temperature to be 61.9 degree C. They then pour 95 mL of hotter water, initially at 85.1 degree C into the calorimeter and measure the temperature of the mixture every 45 seconds over 15-minutes period and obtained a final temperature of 70.2 degree C after plotting a cooling curve. {Assume the density of the water is 2.00 g/mL, and specific heat of water is 6.184 J/g- degree C} Calculate the heat exchanged by the hotter water (q_hw)? Calculate the heat exchanged by the water initially in the calorimeter (q_cw)? Calculate the heat exchange of the calorimeter (q_cal)? Calculate the heat capacity of the calorimeter (C_cal)?Explanation / Answer
From the given data
Heat exchanged by hot water = mCpdT
= 95 x 2 x 6.184 x (85.1 - 70.2)
= 17.507 kJ
Heat exchanged by cold water = mCpdT
= 20 x 2 x 6.184 x (70.2 - 61.9)
= 2.053 kJ
heat absorbed by calorimeter (qcal) = 17.507 - 2.053
= 15.454 kJ
Heat capacity of calorimeter = 15.454/(70.2 - 61.9)
= 1.862 kJ/oC
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