Question

How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol • °C), and the molar heat capacity of ice is 36.4 J/(mol • °C).

Answer 1

Heat released when water vapour is condensed to liquid(q1)

No of mole = 105g/18.013g/mol =5.829

∆H_cond= -40.67kJ/mol

q1 = -40.67kJ/mol × 5.829mol = -237.1kJ

Heat released when liquid water is cooled from 100℃ to 0℃ (q2)

q2 = n × ∆T × C_water

=- (5.829mol × 100℃ × 75.4J/mol ℃)

= - 43.95kJ

Heat released when liquid water freeze to ice(q3)

∆H_freeze = -6.01kJ/mol

q3 =- (6.01 kJ/mol × 5.829 mol =-35.03kJ

Heat released when ice is cooled from 0℃ to -15℃ (q4)

q4 = -(n × ∆T × C_ice)

   _{=- ( 5.829mol × 15℃ × 36.4J/mol℃)}

   _{= -3.18kJ}

_{Total heat release = q1 + q2 + q3 + q4}

   _{= - ( 237.1kJ + 43.95kJ + 35.03 + 3.18)}

   _{= -319.3kJ}

_Therefore,

_{319.3kJ of heat is released}