Question

What amount of ice at `-14^(@)C` required to cool `200gg` of water from `25^(@)C` to 10CC?
(Given `C_("ice")=0.5cal/g.^(@)C,L_(f)` for ice `=80cal//g`)
A. `31g`
B. `41g`
C. `51g`
D. `21g`

Answer 1

(A) Heat given by water in cooling from `25^(@)` to `10^(@)C`
`Q_(1)=(cm DeltaT)_(w)=200xx1xx(25-10)=3000cal`
Heat absorbed by `mg` of ice at `-14^(@)C`
`Q_(2)=(mc DeltaT)_("ice")+mL_(f)+(mcDeltaT)_(w)`
`=m[(0.5)[0-{-14}]+80+1(10-0)]`
`=97mcal`
`Q_(2)=Q_(1)`
`97m=3000=30.93g`
`m=31g`