`1` kg of ice at `0^(@)C` is mixed with `1.5` kg of water at `45^(@)C` [latent heat of fusion = `80` cal//gl. Then

Options

A. the temperature of the mixture is `0_(@)C`
B. mixture contains `156.25` g of ice
C. mixture contains `843.75` g of ice
D. the temperature of the mixture is `15^(@)C`.

Answer

A, B

Explanation

On bringing of `1xx5`kg of water `45^(@)_(c) "to" 0^(@)c` heat released = `1500xx1xx45`cal.
mass of ice that melts with this heat `=(1500xx45)/(80)=843.75`gm
This is less than `1`kg so the resulting temperature is `=0^(@)c`.