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)`)

Options

A. `"458 J Kg"^(-1)K^(-1)`
B. `"916 J Kg"^(-1)K^(-1)`
C. `"654 J Kg"^(-1)K^(-1)`
D. `"1232 J Kg "K^(-1)`

Answer

B

Explanation

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)