Time taken by a $$836$$ W heater to heat one litre of water from $$10^\circ$$C to $$40^\circ$$C is

The total heat energy ($$Q$$) required to raise the temperature of the water is given by:

$$Q = ms\Delta T$$

The heat energy supplied by the heater in time $$t$$ is:

$$Q = P \times t$$

Assuming there is no heat loss to the surroundings, we equate the two expressions for energy:

$$P \times t = ms\Delta T$$

Substituting the given values into the equation:

$$836 \times t = 1 \times 4180 \times 30$$

$$836 \times t = 125400$$

$$t = \frac{125400}{836}$$

$$t = 150 \text{ s}$$