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A 2200-W oven is connected to a 240-V source. R= 26.2 ohm. How long will it take to bring 150 mL of 23 degree C water to 100 degree C assuming 63% efficiency? The specific heat capacity of water is 4186 J/kg*degree C? How much will this cost at 10 cents/kWh?
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Added Fri, 03 Jul '15
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As we know that

Q = msdT/Efficiency

= (150*10^-3)*4186*(100 - 23)/0.63

= 76743.333 J

So

As we know that

Power = Heat Suplied/Time

Thefefore

Time = 76743.333/2200

= 34.883 sec

As we know that

1J = 2.778*10^-7 KWh

Therefore

Cost = 10*(76743.333*2.778*10^-7)

= 0.213 cents
Edit
Added Fri, 03 Jul '15
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