An iron of mass 10 g and of specific heat `"0.1 cal.g"^(-1).^(@)C^(-1)` is heated in a furnace and quickly transferred to a thick - walled copper vessel of mass 200 g and of specific that `"0.09 cal.g"^(-1).^(@)C^(-1)`, kept at `50^(@)C`. The vessel along with its contents, is placed in a calorimeter of water equivalent 20g, containing 180 g water at `20^(@)C`. A thermometer dipped in the water of the calorimeter shows the maximum temperature to be `26^(@)C`. Find the temperature of the furnace. Will there be any local boiling of water in the calorimeter?

Explanation

Let the temperature of the furnace be `theta.`
Heat lost by the hot iron ball and copper vessel
`=10xx0.1xx(theta-26)+200xx0.09xx(50-26)`
`=(theta-26)+18xx24=(theta+406)" cal"`
Heat gained by the calorimeter and water
`=(20+180)xx(26-20)="1200 cal"`
`because" Heat gain = heat loss"`
`therefore" "theta+406=1200" or, "theta=1200-406=794^(@)C`
`therefore" Temperature of the furnace is "794^(@)C`.
Let the intermediate temperature attained by the iron ball and copper vessel bet.
As heat lost by the iron ball = heat gained by the copper vessel
`"So, "10xx0.1xx(794-t)=200xx0.09xx(t-50)`
`"or, "794-t=18t-900`
`therefore" "t=89.16^(@)C`
`therefore" Temperature of the system of iron ball and copper vessel is " 89.16^(@)C`. This is below the boiling point `100^(@)C` of water. So there will be no local boiling in the calorimeter water.