Problem 17.15

17.15. a)
Let's compare the heat required to melt all of the ice with that lost when the water is cooled from $18{}^{\circ}C$ to $0{}^{\circ}C$.
Q_ice = mL
$Q_{ice} = (.250 kg)(3.33 \times 10^{5} J/kg)$
$Q_{ice} = 8.33 \times 10^{4} J$
And the heat lost when the water cools is given by
$Q_{water} = mc\Delta T$
$Q_{water} = (.600 kg)(4184 J/kg)(18{}^{\circ}C)$
$Q_{water} = 4.52 \times 10^{4} J$
So, $4.52 \times 10^{4} J$ goes in to melting the ice and the the system is in equillibrium at $0{}^{\circ}C$.

b)
If $4.52 \times 10^{4} J$ goes in to melting the ice, then the amount of ice melted is given by
$(8.33 \times 10^{4} J - 4.52 \times 10^{4} J) = m(3.33 \times 10^{5} J/kg)$
solving for the mass,
m = .136 kg
So, the amount of ice remaining is
m = .250 kg - .136 kg  =  .114 kg