Two 20.0-g ice cubes at –14.0 °C are placed into 225 g of water at 25.0 °C. Assu

Question

Two 20.0-g ice cubes at –14.0 °C are placed into 225 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts.

Two 20.0-g ice cubes at -14.0 Care placed into 225 g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts. heat capacity of H20(s) 37.7 J/(mol. K) Number heat capacity of H20(D 75.3 J/(mol. K) T 55.2 enthalpy of fusion of H20 6.01 kJ/mol

Explanation / Answer

two 20-g ice cubes = 40-g/18.0-g/mol = 2.22 mol; 225-g/18.0g/mol = 12.5 mol water; one additional point, the heat needed for a 1 K change in temp is the same as for a 1 °C change so the unit J/mol*K will be the same for J/mol*°C

(2.22-mol x 14 °C x 37.7-J/mol*°C) + (2.22-mol x 6010-J/mol) + [2.22-mol x (Tf - 0 °C) x 75.3-J/mol*°C] = 12.5-mol (25 °C - Tf) x 75.3-J/mol*°C

1171.72 + 13, 342.2 + 163.17Tf = 23531 - 941.25Tf ====> (combine like terms)
1104.42Tf = 9017.08
Tf =8.16 °C

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