Question

a water heater marked 2kw and 220v is connected to the main supply. how long will it take to heat 200kg of water from 10°c to 30°c if specific heat of water is 4200J per kg°c?

Answer 1

The total electrical energy supplied to boil water in the kettle isW = VItSince nothing is clearly mentioned about the source of electrical power, we assume that the power source is the same, that means the electrical energy obtained from the power source is the same, that is the current is same.R is the resistance of the coil in the electric kettle.



We have to calculate t.The time t taken to boil the same amount of water for 120 Vsupply is equal toSolving this equation we get , t = 1008.33s = 16.8 minutes.Q)A potential difference of 220 V is applied across a resistance of 1 kW. Calculate the heat energy produced in 10 seconds.Sol:The heat energy produced can be calculated fromH = I2Rt joules.Here I can be calculated from Ohm‘s law V = IR, V–220V and R=1 kW\I = 0.22 AFrom this H = 0.22 x 0.22 x 1000 x 10 = 484 J