A glass beaker contains 200g of carbonated water initially at `20^(@)C`. How much ice should be added to obtain the final temperature of `0^(@)C` with all ice melted, if the initial temperature of ice is `-10^(@)C`. Neglect heat capacity of glass. [Take, `C_("water")= 4190J//kg^(@)C`,
`C_("ice") = 2100J//Kg^(@)C, L_(P) = = 3.34 xx 10^(5)J//Kg`]

Options

A. 47g
B. 76g
C. 200g
D. 22g

Answer

A

Explanation

Heat lost by carbonated water = Heat gained by ice
Let, `m_(i)`= mass of ice melted and `m_(w)` = initial mass of carbonated water.
`m_(w) c_(w) (20 -0) = m_(i) L_(f) + m_(i) (c_(i)) [0 - (-10)]`
`rArr 200 xx 10^(-3) xx 4190 xx 20`
`= m_(i) (3.34 xx 10^(5) + 2100 xx 10)`
`rArr m_(i) = 47 xx 10^(-3) kg or m_(i) = 47g`