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Question

If 21.73g of Al at 398.3 (.C) is added to 151.0g of water at 20.4 (.C), what is the final temperature (.C) of the mixture?

Enthalpy Values : Specific Heat of Al (s)= 0.902 J/g-C
                                 Specific Heat of H20 (l)= 4.184 J/g-C
                                 Qf for H20= 6.01 kj/mol
                                 Gv for H20= 40.67 kj/mol

ChemistryMom · Accepted Answer

so the equation to use is q = mcΔT
assuming no heat loss to the environment, heat gain of water = heat loss of the Al
(21.73 g Al)(0.902 J/gºC)(398.3ºC - Tf) = (151.0 g H2O)(4.184 J/gºC)(Tf - 20.4 ºC)

solving for Tf
Tf = 31.77 ºC

Thermochemistry Problem Need Help!?

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