A piece of iron of mass `100g` is kept inside a furnace for a long time and then put in a calorimeter of water equivalent `10g` containing `240g` of water at `20^(@)C`. The mixture attains an equilibrium temperature of `60^(@)C`. Find the temperature of the furnace.
Specific heat capacity of iron `= 470J kg^(-1).^(@)C^(-1)`

Answer

`[953.6^(@)C]`

Explanation

Here,`m_(1)=100 g, w=10 g, m_(2)=240 g`
`T_(1)=20^(@)C`, Equilibrium temp. `T=60^(@)C`
Let, `T_(2)` be the team. Of furnce.
Heat gained by water and cal. = Heat lost by iron
`(m_(2)+w)(T-T_(1))=m_(1) s(T_(2)-T)`
`(240+10)(60-20)=100xx(470(T_(2)-60))/(4.2xx10^(3))`
`T_(2)-60=(250xx40xx4.2xx10^(3))/(100xx470)=893.6`
`T_(2)=893.6+60=953.6^(@)C`