An unknown metal of mass 192 g heated to a temperature of 100" "^(@)C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4" "^(@)C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5" "^(@)C. (Specific heat of brass is "394" J Kg"^(-1)K^(-1))

An unknown metal of mass 192 g heated to a temperature of 100" "^(@)C

1.	An unknown metal of mass 192 g heated to a temperature of 100" "^(@)C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4" "^(@)C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5" "^(@)C. (Specific heat of brass is "394" J Kg"^(-1)K^(-1))
Answer» `"458 J Kg"^(-1)K^(-1)` `"916 J Kg"^(-1)K^(-1)` `"654 J Kg"^(-1)K^(-1)` `"1232 J Kg "K^(-1)` Solution :Heat LOST by metal piece = heat GAINED by (calorimeter + water) `M_(m)xxC_(m)xx(theta_(2)-theta_(1))=(M_(w)C_(w)+M_(w)xxC_(w))(theta_(2)-theta_(1))`. .(1) Here `M_(m)=` mass of metal piece = 192 g = 0.192 g `M_(w)=` mass of water = 240 g = 0.240 kg `C_(m)=` specific heat of metal = ? `C_(w)=` specific heat of water `=4186" kg"^(-1)K^(-1)` `theta_(2)=` TEMP. of metal piece `=100" "^(@)C` `theta_(1)` = steady temp. of mixture `=21.5" "^(@)C` `theta=` initial temp. of calorimeter + water `8.4" "^(@)C` By SUBSTITUTING values in equation (1), `:.0.192xxC_(m)(100-21.5)=(0.128xx394+0.240xx4186)[21.5-8.4]` `0.192xxC_(m)xx78.5=(50.432+1004.64)xx13.1` `15.072C_(m)=1055.072xx13.1` `15.072C_(m)=13821.4432` `:.C_(m)=(13821.4432)/(15.072)` `=917.0" J kg"^(-1)K^(-1)` `=916" J kg"^(-1)K^(-1)` (nearest value)