Electrical energy. A current I through a wire of resistance R from a supply of voltage V transfers energy at the rate P = VI = V²/R = I²R. In a time t the supply does work

E = V I t = (V²/R) t

In the resistor, the charges collide with the atoms of the metal and give them energy, so the wire gets hot. This is the heating effect of a current (Joule heating).

Heat and temperature. The wire passes the energy to the water. The heat absorbed by water of mass m that warms by ΔT is

Q = m c ΔT, c = 4.18 J/(g °C) for water

where c is the specific heat capacity: the energy needed to raise the temperature of 1 g by 1 °C. Water has a very high specific heat capacity, so it takes a lot of energy to warm it.

Conservation of energy. If all the electrical energy ended up in the water, E = Q. In practice some is lost to the beaker, the air, the wire itself and by evaporation. The energy that went elsewhere is E − Q, and the efficiency is Q/E. The hotter the water is above the room, the faster it loses heat (Newton’s law of cooling), so the heating curve bends over and the efficiency drops. A heater with a low power loses a larger fraction of its energy, because it takes longer, and the jacket reduces the loss. A larger volume of water has a smaller surface area for its mass, so it loses a smaller fraction.

Try it: predict how long it takes for 100 mL to reach 100 °C at 12 V with no loss, then check. Then repeat with heat loss and with the jacket and find the energy that was lost.

The experiment is sped up so that you do not have to wait: at 6 V and 100 mL it would take 37 minutes in real time. The heat loss in the second mode is a simple estimate (0.20 W per °C above the room for 100 mL, halved by the jacket). The heat capacity of the beaker and the wire are ignored. The simulation stops when the water reaches 100 °C.