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)`

Explanation

Let temperature of furnace be `theta`
`m_i=100g`,`m_omega=240g`, water equivalent `m'_omega=10g`.
Equilibrium temperature `=60^@C`
Heat given `=` Heat taken
`m_(i)s_(i)(theta-60)=(m'_(omega)+m_(omega))s_omega(60-20)`
`100xx470(theta-60)=(10+240)xx4200xx40`
`theta-60=(250xx4200xx40)/(100xx470)=893.6`
`theta=953.6^@C`