In the following experiment, a coffee-cup calorimeter containing 100 mL.of H2O i

Question

In the following experiment, a coffee-cup calorimeter containing 100 mL.of H2O is used. The initial temperature of the calorimeter is 23 degree C If 4.10 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution Delta H soln of CaCl2 is -* 82.8 kJ/mol. Assume the density of solution is 1 g/mL. Note: Here Heat of solution Delta H soln is refering to the enthalpy change for dissolving 1 mol of CaCI2(s) in water, or the following process: CaCI2(s) rightarrow CaCI2(aq) Delta H = - 82.8 kJ

Explanation / Answer

4.10 g CaCl2 x (1 mol / 111 g) = 0.0369 mol 0.0369 mol CaCl2 x (82.8 kJ / mol) x (1000 J / kJ) = 3058.4 J Assume that the heat capacity of the solution is the same as that of water (4.184 J/gC) heat = mass x specific heat x delta T delta T = heat divided by (mass x specific heat) delta T = 3058.4 J divided by (100 x 4.184) = 7.3 degrees Final temp should be 23 + 7.3 = 30.3 degrees C

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