When 28.2 mL of 0.500 M H2SO4 is added to 28.2 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction...

Question

(Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.) Do not answer in scientific notation.

Answer 1

q = mC∆T

q = heat of reaction = ∆H

m = mass = 28.2 ml + 28.2 ml = 56.4 ml x 1 g/ml = 56.4 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 30.17º - 23.50º = 6.67º

∆H = (56.4 g)(4.184 J/gº)(6.67º)

∆H = -1574 J (the sign would be negative since the reaction is exothermic)