Question

A 3.56 g sample of solid NH₄C₂H₃O₂ (s) is dissolved in 418 mL of water in a coffee cup calorimeter. Once all of the NH₄C₂H₃O₂ (s) is dissolved in the water, the final temperature of the solution is found to be 24.81°C. If the initial temperature of the water in the calorimeter was 20.81 °C, calculate the calorimeter constant (in J/K) for the coffee cup calorimeter.

Report your answer to three significant figures.



The heat of solvation of NH₄C₂H₃O₂ (s) is -2.38 kJ/mol.

Answer 1

Calorimeter constant of the calorimeter c= q/∆T

Given that heat of solvation of NH4C2H3O2 (s) = -2.38 kJ/mol = - 2380 J/mol

mass of NH4C2H3O2 = 3.56 g

Molar mass of NH4C2H3O2 = 77.0 g/mol

Moles of the compound = mass of compund/ Molar mass of compound

= 3.56 g/ 77.0 g/mol

= 0.046 mol

Hence, Heat released to calorimeter q = ( 2380 J/mol) ( 0.046 mol) = 109.48 J

∆T = 24.81°C - 20.81 °C = 4.0 °C = 4.0 K

Hence, calorimeter constant of the calorimeter

c =  q/ ∆T

= 109.48 J/ 4.0 K

= 27.37 K

c = 27.37 K

Therefore, calorimeter constant of the calorimeter = 27.37 K