Answer:
1.15×10⁶ J
Explanation:
Heat needed to bring the ice to 0°C:
q = mCΔT
q = (3 kg) (2100 J/kg/K) (0°C − (-2°C))
q = 12,600 J
Heat needed to melt the ice:
q = mL
q = (3 kg) (3.36×10⁵ J/kg)
q = 1,008,000 J
Heat needed to bring the water to 10°C:
q = mCΔT
q = (3 kg) (4200 J/kg/K) (10°C − 0°C)
q = 126,000 J
The total heat is:
q = 12,600 J + 1,008,000 J + 126,000 J
q = 1,146,600 J
q = 1.15×10⁶ J
