Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.35 kg of water decreased from 109 °C to 52.0 °C.

Answer 1

You haven't provided any thermodynamic properties, but we will go with generally accepted values.

Csteam = 1.89 J/g/deg

∆Hvap = 2260 J/g

Cliquid = 4.184 J/g/deg

heat liberated when 1.35 kg H2O(g) at 109ºC goes to H2O(g) at 100ºC:

q = mC∆T

q = 1.35 kg x 1000 g/kg x 1.89 J/g/deg x 9 deg = -15,309 J = -15.3 kJ

heat liberated when 1.35 kg H2O(g) at 100ºC condenses to H2O(l):

q = m x ∆Hvap = 1.35 kg x 1000 g/kg x 2260 J/g = -3,051,000 J = -3051 kJ

heat liberated when 1.35 kg H2O(l) at 100ºC cools to 52.0ºC:

q = mC∆T = 1.35 kg x 1000 g/kg x 4.184 J/g/deg x 48ºC = -271,123 J = - 271 kJ

Total energy = -15.3 kJ + -3051 kJ + -271 kJ = -3337 kJ