Answer:
0.12 kg
Step-by-step explanation:
The amount of energy added is:
1000 W × (5 min × 60 s/min) = 300,000 J = 300 kJ
Heat to boil the water is:
q = mCΔT + mL
q = m (CΔT + L)
300 kJ = m (4.2 kJ/kg/K × (100°C − 30°C) + 2200 kJ/kg)
300 kJ = m (2494 kJ/kg)
m = 0.12 kg