In this experiment, you will investigate how much energy an electric kettle element transfers to water. The kettle is rated at 2000 W. You places 500 g of w...

(a) Initial and final temperatures

Initial temperature = 18.0 °C (at 0 s); final temperature = 52.3 °C (at 180 s). [1]

(b) Total temperature rise

\(\Delta T = 52.3 - 18.0 = \mathbf{34.3}\) °C [1]

(c) Energy transferred to the water

Using the thermal energy equation with mass in kg:

\(E = mc\Delta T = 0.500 \times 4200 \times 34.3\) [1]

\(E = \mathbf{72\,030}\) J (approximately 72 000 J) [1]

(d) Energy supplied by the element

\(E = P \times t = 2000 \times 180 = \mathbf{360\,000}\) J [1]

(e) Efficiency of the kettle

Efficiency compares the useful energy output (heating the water) with the total energy input (from the element):

\(\text{Efficiency} = \dfrac{\text{useful energy out}}{\text{total energy in}} \times 100\) [1]

\(\text{Efficiency} = \dfrac{72\,030}{360\,000} \times 100 = \mathbf{20.0\%}\) [1]

This very low efficiency indicates that most of the electrical energy is not being transferred to the water.

(f) Two reasons why efficiency is less than 100%

Any two from: [2]

• Heat is lost to the kettle body and walls, warming the kettle material rather than the water.
• Heat escapes to the surroundings through the opening or lid by convection and radiation.
• Some energy heats the element itself rather than the water.
• Evaporation from the water surface carries energy away as latent heat in the steam.

(g) Improving efficiency

Any one valid method: [1]

• Use a well-fitting lid to reduce heat loss by convection and evaporation.
• Insulate the kettle body to reduce heat conduction through the walls.
• Use a larger mass of water (the percentage of energy lost to the kettle body becomes smaller relative to the energy transferred to the water).

(h) Average rate of temperature rise

\(\text{Rate} = \dfrac{\Delta T}{t} = \dfrac{34.3}{180} = \mathbf{0.19}\) °C per second [1]