heat absorbed by H2O = [mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal x (Tfinal-Tinitial)
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.
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 metal-water 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.
4 answers
5.20
A 94.7-g 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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