A cube of iron (density`= 8000kg m^(-1)`, specific heat capacity `= 470g J kg^(-1)K^(-1)`) is heated to a high temperature and is placed on a larger block of ice at `0^(@)C`. The cube melts the ice below it, displaces the water and sinks in the final equilibrium position its upper surface is just inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube .The density of ice `= 900 kg m^(-1)` and the latent heat of fusion of ice `= 3.36 xx 10^(5)J kg^(-1)`

Answer

C

Explanation

Let the velume of the cube = V
So volume of ice displace = V
Let the initial the tamp be T K
`Mass of cube = 8000V//kg`
We know, beat gained = beat lost
`rArr 8000v xx 470 xx (T - 273)`
`= 900V xx 3.38 xx 10^(8)`
`rArr 376 T - 376 xx 273 = 30420`
`rArrT= (30420 + 102648)/(376)`
`= (133068)/(376)`
` = 353.9042^(@)K`
` = 80^(@)C`