Answer
$10.27^{\circ}C$
Work Step by Step
We first find the value of Q for changing the ice to 0 degrees Celsius liquid:
$Q = m_{ice}c_{ice}\Delta T_{ice}+m_{ice}L_f$
$Q=(.05)(2.05)(10)+(.05)(334)=17.7\ kJ$
We find the energy that would allow the warm water to be cooled to 0 degrees Celsius:
$Q = m_{water}c_{water}\Delta T_{water}=(1)(4.184)(15)=62.8kJ$
Thus, the ice will all melt, and it will heat to:
$T = \frac{62.8-17.7}{4.184\times 1.05 } = 10.27^{\circ}C$
