An aluminium vessel of mass `0.5kg` contains `0.2kg` of water at `20^(@)C` A block of iron of mass `0.2kg at 100^(@)C` is gently put into the water .Find the equilibrium temperature of the mixture, Specific heat capacities of aluminium , iron and water are `910 J kg^(-1)K^(-1) 470J kg^(-1)K^(-1) and 4200J kg^(-1)K^(-1)` respectively

Answer

C

Explanation

Given mass of aluminium `= 0.5 kg`
Mass of water` = 0.2 kg`
Mass of iron` = 0.2 kg`
Tempetature of aluminium and water
`= 20^(@) = 293^(@) K`
Tempetature of iron `= 100^(@)C = 373 k`
Specfic beat of AI ` = 910 J// kg-K`
Heat gain ` = 0.5 xx 610 (T - 293) + 0.2 xx 4200 xx (T - 293)`
`= (T- 293)(0.5 xx 910 + 0.2 xx 4200)`
Heat lost `= 0.2xx 470 xx (373 - T)`
We know heat gain = Head lost
`rArr (T - 293)(0.5 xx 910 + 0.2 xx 4200)`
`= 0.2 xx 470 xx (373 - T)`
`rArr(T - 293)(455 + 840)= 94(373 - T)`
`rArr (T - 293) (1295)/(94) = (273 - T) [(1295)/(94) =- 14]`
` (T - 293) xx 14 = 373 - T`
`rArr 14T - 293 xx 14 = 373 - T`
rArr 15T = 373 + 4102 = 4475`
`rArr T = (4475)/(15) = 298 k `
` :. T = (298 - 273)^(@)C = 25^(@)C`
`:. ` The final temp ` = 25^(@)C`