Question

[References) A 18.0-g sample of ice at -13.6°C is mixed with 105.1 g of water at 80.0°C. Calculate the final temperature of the mixture, assuming no heat loss to the surroundings. The heat capacities of H2O(8) and H2O(l) are 2.03 J/g.°C and 4.18 J/g.°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol. Temperature = ГС Submit Answer Try Another Version 5 Item attempts remaining

Answer 1

Final temperature Tf heat gained by ice- Heat lost by water calculating heat gained by ice mass of ice 18 m g AT ice state from -13.6 °C 0 °C 0 )- -13.6 13.6 to heat capacity of ice H20(s) 2.03 J/g°C heat required qice=m H20(s) AT =18 2.03 13.6 496.9J X X Calculating moles: 18 g mass of ice 18.0 g/mol molar mass of ice moles mass/ molar mass 18/ moles of ice = 1.00 mol 18.0 n Melting of ice: heat required qmelting nAHmelting enthalpy of fusion AHmelting 6020 J/mol kJ/mol 6.02 1.00 x 6020 6020.0 °C Tf °C Tf Tf water state from 0 0 AT = to heat capacity of ice H20() 4.18 J/g°C heat required qwater-m H20(I) AT Tf 75.2Tf 18 x 4.180 heat gained by ice qice +qmelting+water heat gained by ice heat gained by ice 75.2 Tf 497 6020.0 75.2 Tf 6516.9 +

calculating Heat lost by water Mass of water m 105.1 heat capacity of water 4.18 J/g°C Cp = °C Initial temperature =Ti 80.0 °C Final temperature = Tf ? Heat lost by water q m Cp (Ti-Tf) Heat lost by water q 105.1 x 4.18 Heat lost by water q35145 80.0 Tf) 439 Tf calculating final temperature heat gained by ice-Heat lost by water 6516.9 75.2 Tf 35145 439 Tf 514.6 Tf 28628.5 Tf 55.64 °C