200 g of water at `50.5^@C` is cooled down to `10^@C` by adding m g of ice cubes at `0^@C` in it. Find m. Take, specific heat capacity of water = `4.2Jg^(-1)""^@C^(-1)` and specific latent heat of ice = `336Jg^(-1)`

Explanation

Given, ,mass of water = 200 g, intial temperature = `50.5^@C`, final tempreature = `10^@C`
Heat given by water = mass of water `xx` specific heat capacity of water `xx` fail in terperature
`=200gxx4.2Jg^(-1)""^@C^(-1)xx(50.5-10)^@C=34020J" "...(i)`
Heat taken by ice at `0^@C` to melt = mass of ice `xx` specific latent heat of ice
`=mgxx336Jg^(-1)=336mJ`
and heat taken by iced-water at `0^@C` to raise up to `10^@C` = mass `xx` specific heat capacity `xx` rise in temeprature
`=mgxx4.2Jg^(-1)""^@C^(-1)xx(10-0)^@C=42mJ`
`therefore" "` Total heat taken by ice = 336 m+42 m = 378 m J
By the principal of method of mixture
Heat given by water = heat taken by ice
i.e., 34020=378m or `m=(34020)/(378)=90g`