How many grams of ice at ` -14 .^(@)C` are needed to cool 200 gram of water form `25 .^(@)C` to `10 .^(@)C`? Take specific heat of ice `=0.5 cal g^(-1) .^(@)C^(-1)` and latant heat of ice = `80 cal g^(-1)` .

Options

A. 11g
B. 21g
C. 31g
D. 41g

Answer

C

Explanation

In cooling 200g of water from `25^(@)C " to " 10^(@)C` heat to be extracted from water is
`Q_(1) = m_("water") s_("water") (25-10) = 200 xx 1 xx (25-10) = 3000` cal
Heat absorbed by `m_("ice")`g ice at `-14^(@)C` to convert it to water at `10^(@)C` is
`Q_(2) = m_("ice") s_("ice") (0-(-14) + m_("ice") L_(f_("ice")) + m_("ice") s_("water") (10-0)`
`=m_("ice") xx 0.5 xx 14 + m_("ice") xx 80 + m_("ice") xx 1 xx 10 = 97m_("ice")`
As `Q_(1) = Q_(2)`
`:. 97m_("ice") = 3000 or m_("ice") = 31g`