3.Why is ΔTwater > 0 ? Because the water gains heat from the metal. 4.A metal sample weighing 43.5 g and at a temperature of 100.0 °C was placed in 39.9 g of water in a calorimeter at 25.1 °C. At equilibrium the temperature of the water and metal was 33.5 °C. Determine the specific heat capacity of the metal.
5.What is the average specific heat capacity of the unknown metal in this experiment?a.Q=mc(DeltaT)Cmetal= Qmetal/( Mmetal*DeltaT Metal)b.Mass of Metal = 19.7g, DeltaTmetal= the final temp of the water- 100 degrees Celsius. (boiling water) c.Qmetal= Qwaterd.Trial 1: (50gH2O)(4.184 J/g)(30-6C)= -836.8J -836/ (-70*19.7) = .607J/gCe.Trial 2: (50gH2O)(4.184 J/g)(31-26C)=-1046J -1046/(-69*19.7)= .76 J/gCf.Trial 3: (50gH2O)(4.184 J/g)(32-27C)=-1046 -1046/(-68*19.7)= .78 J/gCg.Average: .72 J/gC6.What is the unknown metal? Use Table 1 for reference.
Enthalpy and Specific Heat