Answer
$19.1C^{\circ}$
Work Step by Step
We know that the heat lost by the lemonade is given as
$Q_w=m_{lem}C_{water}\Delta T$
$Q_w=(3.99Kg)(4186J/Kg.C^{\circ})(20.5C^{\circ}-T_f)$
$\implies Q_w=16702-14(20.5C^{\circ}-T_f)J$
The heat gained in raising the temperature of water to $T_f$ is given as
$=m_{ice}C_{ice}\Delta T+m_{ice}L_f+m_{ice}C_{water}\Delta T$
$=(0.055Kg)(2090J/KgC^{\circ})(0C^{\circ}-(-10.2C^{\circ}))=19597.5+230.23(T_f)$
Now, according to the principle of calorimetry:
heat gained by ice=heat gained by lemonade
$Q_{ice}=Q_w$
$\implies 19597.5+230.23(T_f)=16702(20.5C^{\circ}-T_f)J=19.1C^{\circ}$
