A piece of iron of mass `100g` is kept inside a furnace for a long time 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 t.
Given, `m_(1) = 100g, m_(w) = 240g`,
water equivalent `m_(w)^(')`=10g,
Equilibrium temperature = `60^(@)`C.
Then from principle of calorimetery,
Heat given = Heat taken
`m_(i)s_(i)(t-60) = (m_(w)^(')+m_(w))s_(w)(60-20)`
`100xx470(t-60) = (10 + 240) xx 4200 xx 40`
`(t-60) = (250xx 420 xx40)/(100 xx 470) = 893.6`
`t = 953.6^(@)C`