Question

From the data below, calculate the total heat (in J) needed to convert 15.00 g of ice at −8.75°C to liquid water at 0.650°C: m.p. at 1 atm: 0.0°C ΔH o fus : 6.02 kJ/mol c liquid: 4.184 J/g·°C csolid: 2.09 J/g·°C

Answer 1

Mass of ice = 15 g Number of moles of ice = 15 g/(18.015 g/mol) = 0.832639 mol AHfus = 6.02 kJ/mol Specific heat capacity of ice (solid) = 2.09 J/(g. °C) Specific heat capacity of water = 4.184 J/(g. °C) This process involves steps. 1. Heat associated with the conversion of ice at -8.75 °C to ice at 0 °C 91 = m*C*AT = 15 g * 2.09 J/(g.°C)* (0 - (-8.75 ) )°C = 274.3125 J 2. Heat associated with the conversion of ice at 0°C to water at 0°C. q2 = m* AHfus = 0.832639 mol * 6.02 kJ/mol = 5.01249 kJ = 5012.49 J 3. Heat associated with the conversion of water at 0°C to water at 0.650 °C q3 = 15 g *C*AT = 15 g * 4.184 J/(g. °C)* (0.650 - 0)°C = 40.794 J Total heat energy required = 274.3125 + 5012.49 + 40.794 = 5327.596 kJ = 5327.6 kJ