chemistry
posted by marcus .
A 94.7 g sample of silver (s= 0.237 J/(g x Degree C)), initially at 348.25 Degrees C , is added to an insulated vessel containing 143.6 g of water (s=4.18 J/(g x Degree C)), initially at 13.97 Degrees C. At equilibrium, the finial temperature of the metalwater mixture is 22.63 Degrees C. How much heat was absorbed by the water? The heat capacity of the vessel is 0.244 kJ/Degrees C.

heat absorbed by H2O = [mass H2O x specific heat H2O x (TfinalTinitial)] + [Ccal x (TfinalTinitial)
Take note the Ccal is given in kJ/C and specific heat H2O is given in J/C. You must change one of them; the easier one is Ccal. 
5.20

A 94.7g sample of silver (s = 0.237 J/(g · °C)), initially at 348.25°C, is added to an insulated vessel containing 143.6 g of water (s = 4.18 J/(g · °C)), initially at 13.97°C. At equilibrium, the final temperature of the metal–water mixture is 22.63°C. How much heat was absorbed by the water? The heat capacity of the vessel is 0.244 kJ/°C.

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