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 420J kg^(-1)K^(-1)` respectively.

Explanation

Here container and water are at `20^(@)C` thus when iron blockis dropped into water it will loose energy to tit and its temperature will fall. If equolibrium temperature is `T_(0)`, then we have
heat lost by iron block = heat gained by water plus container
`m_(i)xxs_(i)(100-T_(0))=m_(omega)xxs_(omega)(T_(0)-20)+m_(c)+s_(c)(T_(0)-20)`
`=(0.2xx470xx100+0.2xx4200xx20+0.5xx910xx20)/(0.2xx470+0.2xx4200+0.5xx910)`
`=(9400+16800+9100)/(94+940+455)`
`=35300/1329=26.56^(@)C`.