suppose you want to cool 0.25 kg of cola (mostly water ), at `25^(@)C` by adding ice initially at `-20^(@)C`. How much ice should you add so that the final temperature will be `0^(@)C` with all the ice melt? Neglect the heat capacity of the container. specific heat of ice is` 2000 Jkg^(-1)K^(-1)`. [take specific heat of cola `4160 Jkg^(-1)K^(-1)`.]

Explanation

Heat lost by cola is
`Q_(cola)=m_(cola)c_(cola)DeltaT_(cola)`
`(0.25kg)(4160 Jkg^(-1)K^(-1)25^(@)C - 0^(@)`C ltbrge26000J
Let the mass of the required ice be `m_(ice)`, then the heat needed to warm it from `-20^(@)`C to `0^(@)`C is
`Q_(1) = m_(ice)c_(ice)DeltaT_(ice)`
`m_(ice)(2000)Jkg^(-1)K^(-1))(0^(@)C-(-20^(@)C))`
`m_(ice)(3.34 xx 10^(5)Jkg^(-1)`
`therefore` Total energy gained by ice is
`Q_(1)+Q_(2)=m_(ice)(3.74 xx 10^(5)Jkg^(-1)`
Using principle of calorimetry, heat lost = heat gained, we get
26000 J = `m_(ice)(3.74 xx 10^(5)Jkg^(-1)` or, `m_(ice)`=0.07kg = 70g