Two 20.0-g ice cubes at -17.0 degree C are placed into 225 g of water at 25.0 de

Question

Two 20.0-g ice cubes at -17.0 degree C are placed into 225 g of water at 25.0 degree C. Assuming no energy is transient to or from the surroundings, calculate the final temperature. T_f. of the water after all the ice melts. The heat energy, q. associated with a temperature change. AT, is related to the number of motes, n. and heat capacity. C. of the substance according to the equation for a phase change, the heal energy is related to the number to moles and latent heat of the phase change, which, in this case, is the enthalpy of fusion. Delta H_rms by the equation

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 17 °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

1422.79 + 13, 342.2 + 163.17Tf = 23531.25 - 941.25Tf ====> (combine like terms)
1104.42Tf = 8766.26
Tf =7.94 °C

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