Question

1. A volume of water was heated to 83.90 °C and immediately added to 48.11 mL of water at 23.98 °C contained within a coffee cup calorimeter. The final temperature of the mixture was 40.86 °C. The final volume of water inside the calorimeter was 90.45 mL. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following: a. The volume of hot water added ml b. The mass of hot water (dwater = 1.00 g/mL) c. The mass of the cold water (dwater = 1.00 g/mL) d. The change in temperature (AT) of hot water °C e. The change in temperature (AT) of cold water f. The calorimeter constant in J/°C (C (water) = 4.18 J/g/°C) J/°C

Answer 1

Q1. (a.) volume of hot water added = (total volume after mixing) - (volume of cold water)

volume of hot water added = (90.45 mL) - (48.11 mL)

volume of hot water added = 42.34 mL

(b.) mass of hot water = (volume of hot water added) * (density of water)

mass of hot water = (42.34 mL) * (1.00 g/mL)

mass of hot water = 42.34 g

(c.) mass of cold water = (volume of cold water added) * (density of water)

mass of cold water = (48.11 mL) * (1.00 g/mL)

mass of cold water = 48.11 g

(d.) change in temperature of hot water = (final temperature) - (initial temperature of hot water)

change in temperature of hot water = (40.86 ^oC) - (83.90 ^oC)

change in temperature of hot water = -43.04 ^oC

(e.) change in temperature of cold water = (final temperature) - (initial temperature of cold water)

change in temperature of cold water = (40.86 ^oC) - (23.98 ^oC)

change in temperature of cold water = 16.88 ^oC

(f.) Heat gained by cold water = (mass of cold water) * (specific heat water) * (increase in temperature of cold water)

Heat gained by cold water = (48.11 g) * (4.18 J/g.^oC) * (16.88 ^oC)

Heat gained by cold water = 3394.5646 J

Heat lost by hot water = 7617.2708 J

Heat gained by calorimeter = (Heat lost by hot water) - (Heat gained by cold water)

Heat gained by calorimeter = (7617.2708 J) - (3394.5646 J)

Heat gained by calorimeter = 4222.7 J

calorimeter constant = (Heat gained by calorimeter) / (change in temperature of cold water)

calorimeter constant = (4222.7 J) / (16.88 ^oC)

calorimeter constant = 250 J/^oC

Q2. (a.) The total mass of aqueous solution inside the calorimeter = 94.49 g

(b.) The change in temperature of the aqueous solution = 6.17 ^oC

(c.) The heat released by the neutralization reaction = 2598 J

(d.) The moles of HCl reacted = 0.04633 mol

(e.) The enthalpy change for the neutralization = -56.1 kJ/mol HCl