Question

125 kg. of water is to have it's temperature raised from 18°C to boiling point in 24 minutes. The efficiency of heat transfer is 82%. How much current will be drawn if the heating element is 120? How much will it cost to heat this water if energy costs 10.5ç/kW-h?

Answer 1

So Now 12,500 KJ of energy is consumed in 24 minutes so

Power consumed = 12500 / 24 x 605 = 8.68 k J/s

But 1w = 1J/_Is

So power consumed = 8.68 KW.

* '' Kwatt hour '' is the energy consumed when one watt is used in one hour"

But in our case 8.68 KW is consumed in 24 minutes i.e 24 / 60 hour's = 0.4 hours.

So Now the Total energy consumed will be 8.68 kw x 0.4 hours = 3.47 kwhr given that it will be costing 10.5 kh - h

So the total cost will be = 34.7  

Now

But the Device is Run for 24 minutes only so current consumed = 0.015 x 0.4 = 0.006 mA

--> considering A general Supply Voltage of " 230 v "