Test the efficiency of an immersion heater to determine its cost effectiveness when compared to other heating devices. Immersion heaters consist of an approximately 25-cm (10-inch) long metal bar connected to a power pack that plugs into a wall socket. When plugged in, the metal bar warms up and transfers heat energy, in joules, to the liquid in which it is submerged. Since the transfer of energy is not perfect, the immersion heater requires more energy to operate than it can transfer to the liquid, based on its efficiency. Efficiencies are expressed as percentages.

Determine the mass of a drinking glass in grams. For example, the mass might be 200.0g. Fill the drinking glass with water. Then determine the total mass of the water and glass together. Also, measure the temperature of the water. The combined mass might be 450.0g, and the initial water temperature might be 20.0 degrees.

Subtract the mass of the drinking glass from the total mass to obtain the mass of the water. Performing this step leads to 450.0g minus 200.0g, or 250.0g.

Plug a watt meter into an outlet, and then turn it on. A watt meter measures the power consumption of a device in watts, or joules of energy used per second. Plug the immersion heater into the watt meter, turn it on, and then submerge the heater in the water. Start the stopwatch, and note the reading on the watt meter. Assume, for this example, that the reading is 400.0 watts

Stop the stopwatch when the water in the drinking glass reaches 40.0 degrees Celsius. Assume, for this example, that the time required was 15.0 minutes, or 900.0 seconds.

Subtract the final water temperature from its initial temperature to get the water temperature change. Performing this step yields 40.0 minus 20.0 degrees for a temperature change of 20.0 degrees. Convert the temperature change to Kelvin by adding 273.15. This temperature change is equivalent to 293.15 kelvin.

Multiply the mass of the water times the specific heat of water times the temperature change, to obtain the energy required to heat the water in joules. The specific heat of water, 4.186 joules per gram per Kelvin, is the energy in joules needed to raise the temperature of one gram of water one degree Kelvin. Continuing the exercise, you have 250.0 grams times 4.186 joules per gram per Kelvin, times 293.15 Kelvin, or 306,781.5 joules.

Multiply the watt usage of the immersion heater by the time it was used to arrive at the energy it consumed in joules. Now you have 400.0 joules per second times 900.0 seconds, or 360,000 joules.

Divide the energy required for heating the water by the actual energy consumed by the immersion heater, and then multiply by 100 to get its efficiency as a percentage. Completing the exercise, you have 306,781.5 joules divided by 360,000 joules, or 0.85. The efficiency is therefore 85 per cent.

#### Tip

Use metric units for measurements and calculations, since they are required for the scientific equation used.