I was very mulberry heart when I saw the answer upstairs. Obviously, you can use junior high school physics knowledge to answer! As follows:

First of all, we should know that the specific heat capacity of water is 4.2× 103j/ Celsius (the energy required to raise the temperature once per liter of water).

Suppose that the water at 20℃ rises to 100℃, and the required energy is (100-20) × 4.2×103 = 336000 J.

Because one kilowatt-hour (that is, we often say one degree of electricity) is equal to 3.6 × 10 6 J.

After conversion, the heat dissipation of about 0. 1 kwh accounts for 70 0. 1/0.3=0.3 kwh.

If the volume is 2L, it needs 0.6 kWh of electricity.