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)`

Options

A. `750^(@)C`
B. `101.5^(@)C`
C. `850^(@)C`
D. `953.6^(@)C`

Answer

D

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`