You are confusing units, which is making your question difficult to understand.
Electrical energy is measured in kW-hours (or Joules), not kW. If your heater consumed 500 W of power for 610 s, then you used 305 kJ of energy to heat your water.
Similarly, if 10 l (10 kg) of water was raised by 24.2 °C, then it absorbed
$$4.186 \frac{J}{g ^\circ C} \cdot 10 kg \cdot 24.2 ^\circ C = 1013 kJ$$
of energy. Clearly, something is wrong with your description of what actually happened.
If the heater actually used 0.5 kWh of energy (i.e., 2.95 kW for 610 seconds), this corresponds to 1800 kJ, which would indicate that the process is only about 1013/1800 = 56.3% efficient. You lost quite a lot of the incoming energy in the form of heat that ended up somewhere other than in the water.