Question

028. An electric rod of 2000 watts rating bolls a certain quantity of water
In 10 minutes, the heat which is generated for boiling this water is:
A. 8 x 104 Joules
12 x 105 Joules
19 x 10% Joules
23 x 105 Joules
37 x 105 Joules

Answer 1

Explanation:

Assuming your electric rod is 100% efficient I.e. delivered the exact power it is rated for and putting values in the equation that reads:

Power = Energy/time

Heat energy = Power * Time

Power = 2000 W

Time is 10 minutes = 10 * 60s = 600s

Energy = 2000 W or J/s * 600 s

Heat Energy = 1200000 J or 1200 kiloJoule(kJ) or 1.2 MJ(MegaJoule)

Answer 2

Answer:

So, the answer is: 2 x 10/60 kilowatt-hours = 20/60 kilowatt-hours. That's 1/3 of a kilowatt-hour. That's the amount of energy used, without going into some of the complications mentioned in other answers.

Now let's put that in terms of the usual unit of energy for heat, the joule. One watt is one joule per second. 2000 watts for 10 minutes (600 seconds) is 1,200,000 joules.

That's 1200 kilojoules. So my answer is: 1200 kJ.

Explanation:

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