Question

What amount of ice must be added to 250 g of water at 15°c to cool the water to 0°c and have no ice left

Answer 1

`\mathbf{m= 47 \ g}`

Step-by-step explanation:

The heat needed to melt ice
`(q_1)` + heat needed to cool water
`(q_2)` = 0

`m\Delta H_(fusion) + mc \Delta T = 0`

`m * 333.55 J g^(-1) + 250 \ g * 4.184 \ Jg^(-1 \ 0 ) * (-15.0 \ ^0 C) = 0`

`m * 333.55 J g^(-1) =15690`

`m=(15690 \ J )/(333.55 J g^(-1) )`

`\mathbf{m= 47 \ g}`