`Ice at -20^(@)C` is added to 50 g of water at `40 ^(2)C` When the temperature of the mixture reaches `0^(@) C` it is found that 20 g of ice is still unmelted .The amount of ice added to (Specific heat of water `=4.2 j//g//^(@)C`
Specific heat of Ice `=2.1 J//g //^(@)C` M
Heat of fusion of water of `0^(@)C=334 J//g)`

Options

A. 40 g
B. 60 g
C. 50 g
D. 100 g

Answer

A

Explanation

Heat needed to bring 50 gm water from `40^@C` to `0^@C`
`Q_1=50xx4.2xx40`=8400 J
Out of this some heat is provided by 20 gm ice .
`Q_2=20xx2.1xx20`=840 J
`Q=Q_1-Q_2`=7560
If mass 'm' of ice is melted, then `7560=mxx334+m xx2.1xx20 ` gives ` m approx 20` gm .
so, m(total)=40 gm .