How long will it take to heat 4 kg of water from 30ºC to 65ºC using an electric kettle taking 5 A from a 240 V supply? (Specific heat capacity of water = 42...

Answer Details

This question links the electrical energy supplied by the kettle to the heat energy gained by the water. Assuming no heat is lost, the electrical energy delivered in time \(t\) equals the heat needed to raise the water's temperature:

\[IVt = mc\,\Delta\theta.\]

Work out each side separately. The heat required is

\[mc\,\Delta\theta = 4\times 4200\times (65-30) = 4\times 4200\times 35 = 588\,000\ \text{J}.\]

The power of the kettle is

\[P = IV = 5\times 240 = 1200\ \text{W}.\]

Since power is energy per second, the time taken is

\[t = \frac{588\,000}{1200} = 490\ \text{s}.\]

Two slips account for the other figures. Using the final temperature \(65\ ^\circ\text{C}\) instead of the temperature rise of \(35\ \text{K}\) inflates the energy badly, and halving or doubling the power (for instance by dividing by \(2400\) instead of \(1200\)) gives \(245\ \text{s}\), which is the trap set here. Also note that a temperature change of \(35\ ^\circ\text{C}\) is numerically identical to \(35\ \text{K}\), so the specific heat capacity in \(\text{J kg}^{-1}\text{K}^{-1}\) can be used directly without converting to kelvin. Always compute the temperature difference first and write it down before substituting.