Heat available from water cooling 25°C to 0°C: Qavailable=100×4200×25=10.5×106 J.
Heat to warm ice from -10°C to 0°C: Q1=10×2100×10=0.21×106 J.
Heat to melt ice at 0°C: Q2=10×3.36×105=3.36×106 J.
Total needed: 0.21+3.36=3.57×106 J.
Since available heat exceeds required heat, all ice melts. Remaining heat: 10.5−3.57=6.93×106 J
warms 110 kg water: ΔT=110×42006.93×106=15 °C.
Final temperature: 15°C.
Decrement in water temperature: 25−15=10 °C.