At equilibrium, both metal and calorimeter have the same temperature. By conservation of energy, the energy lost by the metal is gained by the calorimeter:
[tex]heat\,gain\,by\,calorimeter=heat\,loss\,by\,metal[/tex]
[tex]250.0g\times 1.035cal/g/C \times (11.08-10)C = 50.00g\times C_{v}\times (45-11.08)C[/tex]
Where [tex]C_{v}[/tex] is the specific heat of the unknown metal. Solving this gives us the specific heat of the metal:
[tex]C_{v}=0.165cal/g/C[/tex]
