Question

water has a molar heat capacity of 75.38 J/(mol

12. Water has a molar heat capacity of 75.38 J/(mol x °C) and its vaporization enthalpy at 100°C is 40.7 kJ/mol. How much energy is needed to convert 36 g liquid H20 at 70.0°C to steam at 100°C? A) 85.9 kJ B) 126 kJ C) 77.6 kJ D) 81.4 kJ E) 45.2 kJ

Answer 1

Motor heat capacity (c) = 75.38 j/mol - degree C delta H_vap = 40.7 kJ/mol at 100 degree C Energy needed to convert 36 gram liquid H_2 O at 70 degree C to steam at 100 degree C = E This process will have two steps: (i) conversion of 36 gram (liquid) H_2 O at 70 degree C to 100 degree C liquid (ii) conversion of 100 degree C liquid to 100 degree C steam. Thus E = E_1 + E_2 = (moles of H_2 O) times (T_2^uparrow 100 degree C - T_1^uparrow 70 degree C) + delta H_vap times moles of H_2 O = 36gram/(10 grams/mole) times 75.38 J/mol degree times 30 degree + 40.7 times 10^3 J/mol times 36 gram/18 gram/mol E = 60 times 75.38 J + 40.7 times 10^3 times 2 J E = 85.923 kJ Thus option A is correct