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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» Correct Answer - `[953.6^(@)C]` 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` |
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