Question

a reaction was conducted in an "open" (constant pressure) calorimeter. 1.00g acetic acid (HC2H3O2) is dissolved in 50.0 mL h20 (assume density 1.00g/ml specific heat 4.184j/g deg C) the temp of water was initially 25 deg C and final temp was 25.51 deg C. The process being studied is HC2H3O2(l) --> HC2H3O2 (aq). Delta H for hydration of one mole of HC2H3O2 is -7.19KJ. Calculate the Heat Capacity of calorimeter

Answer 1

We should calculate the heat of the calorimeter by taking into account the heat generated by change in temperature and by the calorimeter:

Q = m_solCp_waterdeltaT + Cp_CaldeltaT

We obtain every component we can:

m_sol = 51 mL * (1g/mL) = 51 g

Cp_water = 4.184 J/g C

deltaT = 25.51 - 25 = 0.51 ºC

Q_gen = 7.19 kJ / mol * (1g * (1mol / 60.05g)) = 119.7335 J

Having these data, we substitute and solve for Cp_Cal

119.7335 J = (51g * 4.184 J/gC * 0.51C) + (Cp_cal * 0.51)

119.7335 = 108.8258 + 0.51Cp_cal

10.9077 = 0.51Cp_cal

Cp_cal = 21.3876 J/C