Question

if 22.0 mL of water containing 0.030 mol of HCl is mixed with 38.0 mL of water containing 0.030 mol of NaOH in a calorimeter such that the initial tempature of each solution was 27.0 degrees C and the final temperature of the mixture is 35.0 degrees C, how much heat (in kJ) is realeased in the reaction? Assume that the densities of the solutions are 1.00 g/mL.

Answer 1

Assuming there is no heat loss in calorimeter and specific heat of mixture is same as water specific heat c = 4.186 J/g degree C Amount of heat released by reaction is used to increased is used to increase temperature of solution. Amount of heat released (Q) = Amount of heat absorbed by solution = m_sol^n c_sol^n (T_final - T_) mass of solution = Volume of solu^n times density = (22.0 mL + 38.0 mL) times 1.00 g/mL = 60.0 mL times 1.00 g/mL = 60.0g Specific heat of solution C_ = 4.186 J/g degree C Final temp = 35 degree C Initial temp = 27 degree C Q = (60.0 g) (4.186 J/g degree C) (35 - 27) degree C Q = 2009.28 J Q = 2.009 KJ 2.01 KJ heat is released.