The temperature of 170 g of water at `50^@`C is lowered to `5^@`C by adding certain amount of ice to it. Find the mass of ice added. Given : Specific heat capacity of water = 4200 J `kg^(-1)^@C^(-1)` and specific latent heat of ice = 336000 J `kg^(-1)`.

Explanation

Given : mass of water m =170 g = 1.17 kg, initial temperature `= 50^@C`, final temperature = `5^@`C
Fall in temperature `= Delta t = (50 – 5) = 45^@ C = 45 K `
Heat lost by water `= mc Delta t`
`= 0.17 xx 4200 xx 45`
`= 3.213 xx 10^4 J`
If m. kg ice is added, heat gained by it to melt to `0^@ C = m.L`
`= m. xx 3.36 xx 10^5 `
Heat gained by it to raise temperature by `5^@ C = m.C Delta t `
`= m. xx 4200 xx 5`
` = m. xx 2.1 xx 10^4 J`
Total heat gained by ice `= 3.36 xx 10^5 m. + 2.1 xx 10^4 m`
`= 3.57 xx 10^5 m. J `
By the principle of method of mixtures
heat lost by water = heat gained by ice
`3.213 xx 10^4 = 3.57 xx 10^5 m.`
` m.=( 3.213 xx 10^4)/(3.57 xx 10^5)`
`=0.09 kg ( 90 kg)`