50.00 mL of 1.0 M NaOH solution at 18.0°C is mixed with 50.00 mL of 1.0 M HCI so
ID: 1057051 • Letter: 5
Question
50.00 mL of 1.0 M NaOH solution at 18.0°C is mixed with 50.00 mL of 1.0 M HCI solution also at 18.0°C in a calorimeter. The mixture was allowed to react to completion and the temperature of the mixture rose to 23.60°C. Use the following information to calculate the amount of heat absorbed by the mixture in kJ. Density of the mixture is 1.1 g/mL, Specific heat capacity of mixture 4.18 J/gK. Heat capacity of calorimeter 8.65 J/K. Hint: qreaction= -qsolution. (Note: Express your answer to 2 significant figures and do not include units (KJ) in the calculated answer).
Explanation / Answer
volume of solution = 50 + 50 = 100 mL
denisty of solution = 1.1 g/ml
mass of solution = 100 x 1.1 = 110 g
specific heat = 4.18 j/goC
temperature difference = dT = 23.60 - 18 = 5.60oC
heat Q = m Cp dT + Cp dT
= 110 x 4.18 x 5.6 + 8.65 x 5.60
= 2623 J
amount of heat absorbed = 2.6 kJ
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