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 furnce specfic beat capacity of iron `= 470J kg^(-1)^(@)C^(-1)`

Answer

C

Explanation

Given Mass of iron ` = 100g`
Water equivalent of calory meter
`= 10 g`
Mass of water `= 240 gm`
Let the of surface `= 0^(@)C`
`S_(min) = 470 J// kg`
Tatal beat gained = Total beat lost
`So, (100)/(1000) xx 470 xx (theta - 60^(@))`
`= ((240 + 10))/(1000) xx 4200 xx (60 - 20)`
`rArr 47 theta - 47 xx 60 = 25 xx 42 xx 40`
`rArr theta = (24000 + 2820)/(47)`
` = (44820)/(47) = 953.61 ^(@)C`