18.1 g of water (initially at 20.0 degreeC) is poured onto 270. g of hot, alumin

Question

18.1 g of water (initially at 20.0 degreeC) is poured onto 270. g of hot, aluminum metal (initially at 500.0 degree C). Once all of the water has vaporized (and no heat is lost to the surrounding air), what will be the final temperature of the aluminum? T_mp, al = 660. degreeC C_solid al = 24.2 J/mol degreeC C-m, liquid al= 29.3 J/mol.degee C C_m, liquid water = 75.3 J/mol degreeC, C_m, water vapour= 33.1 J/mol-degreC = 6.02 kJ/mol = 40.67 kj/mol Consider pure substances composed of the following molecules. For each set, put the molecules in

Explanation / Answer

We make an energy balance:

Qgained = -Qlost

Aluminum will just cool, while water will vaporize so:

Qlost = mCpdeltaT

Qlost = 270 g * (1mol/27g) * (24.2 J / molºC) * (Tf - 500)ºC

Qgained = m * [CpdeltaT + deltaHvap]

Qgained = [18.1 g * (1 mol / 18g)] * [(75.3 J / molºC * (100 - 20)ºC) + (40670 J/mol)]

Qlost = 242 Tf - 121000

Qgained = 46953.41

Equalizing:

-242 Tf + 121000 = 46953.41

-242 Tf = -74046.59

Tf = 305.9776 ºC

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