Question

The heater element of an electric kettle has a constant resistance of 100 Ω and the applied voltage is 250 V. Calculate the time taken to raise the temperature of one litre of water from 15ºC to 90ºC assuming that 85% of the power input to the kettle is usefully employed. If the water equivalent of the kettle is 100 g, find how long will it take to raise a second litre of water through the same temperature range immediately after the first.

Answer 1

Mass of water = 1000 g = 1 kg 

(∵ 1 cm³ weight 1 gram) 

Heat taken by water = 1 × (90 − 15) = 75 kcal 

Heat taken by the kettle = 0.1 × (90 − 15) = 7.5 kcal 

Total heat taken = 75 + 7.5 = 82.5 kcal 

Heat produced electrically H = I²Rt/J kcal 

Now, I = 250/100 = 2/5 A, J = 4,200 J/kcal; 

H = 2.52 × 100 × t/4200 kcal 

Heat actually utilized for heating one litre of water and kettle 

= 0.85 × 2.5² × 100 × t/4,200 kcal

∴ (0.85 x 6.25 x 100 x t)/4,200 = 82.5 

∴ t = 10 min 52 second

In the second case, heat would be required only for heating the water because kettle would be already hot.

∴ 75 = (0.85 x 6.25 x 100 x t)/4,200

∴ t = 9 min 53 second