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 :
For hot body :
170 g water at `50^(@)C` changes to water at `5^(@)C`.
`m=170g=170/1000kg,c=4200Jkg^(-1)""^(@)C^(-1),`
`DeltaT=(50-5)^(@)C=45^(@)C`
`therefore` Heat lost by water = `mcDeltaT`
`=170/1000xx4200xx45J`
= 32130 J
For cold body :
Let mass of ice be x kg.
x kg ice at `0^(@)C` changes to water at `5^(@)C`.
Mass (m) = x kg, L = 336000 `Jkg^(-1)`,
`c=4200Jkg^(-1)""^(@)C^(-1),DeltaT=(5-0)^(@)C=5^(@)C`
`therefore` Heat gained by ice = `mL+mcDeltaT`
= `(x xx336000+x xx4200xx5)J`
= (336000 x + 21000 x)J
= 357000 xJ.
When no heat energy is lost to the surroundings,
Heat gained = Heat lost
or 357000 x = 32130
`thereforex=32130/357000kg`
= `0.09kg=90g`.
`therefore` Mass of ice added = 90g.