Question

An electric kettle needs six minutes to boil 2 kg of water from the initial temperature of 20ºC. The cost of electrical energy required for this operation is 12 paise, the rate being 40 paise per kWh. Find the kW-rating and the overall efficiency of the kettle.

Answer 1

 Input energy to the kettle = 12 paise/40 paise/kWh =  0.3 kWh

Input power = energy in kWh/Time in hours = 0.3/(6/60) = 3 kW

Hence, the power rating of the electric kettle is 3 kW 

Energy utilised in heating the water 

= mst = 2 × 1 × (100 − 20) = 160 kcal = 160 /860 kWh = 0.186 kWh. 

Efficiency = output/input = 0.186/0.3 = 0.62 = 62%.