Answer
.176 kg, or 176 grams
Work Step by Step
We first find the value of Q for changing the ice to 0 degrees Celsius liquid:
$Q = m_{ice}c_{ice}\Delta T_{ice}+m_{ice}L_f$
$Q=(.2)(2.05)(10)+(.2)(334)=4.1+66.8=70.9\ kJ$
We find the energy that would allow the warm water to be cooled to 0 degrees Celsius:
$Q = m_{water}c_{water}\Delta T_{water}=(1)(4.184)(15)=62.8kJ$
We find that $62.8-4.1=58.7 \ kJ$ goes to getting the ice to 0 degrees Celsius. This means that:
$m_{melted}=\frac{58.7}{334}=.176\ kg$
.176 kg, or 176 grams
