`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=8400J`
Out of this some heat is provided by 20gm ice.
`Q_(2)=20xx2.1xx20=840J`
`Q=Q_(1)-Q_(2)=7560`
If mass ‘m’ of ice is melted, then `7560=mxx334+mxx2.1xx20" gives "~=20gm`.
So, m(total)=40gm