A 6000 Kg mass of water is brought to a uniform temperature of 98 ˚C in a heavily insulated tank using an electrical heater. The initial uniform temperature of the water is 20 ˚C. The heat loss through the insulation is 5(108) J as the mass of water is heated. The specific heat of water is 4,120 J/(Kg K). Calculate the electrical energy required to perform this process.

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Asked Oct 26, 2019
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A 6000 Kg mass of water is brought to a uniform temperature of 98 ˚C in a heavily insulated tank using an electrical heater. The initial uniform temperature of the water is 20 ˚C. The heat loss through the insulation is 5(108) J as the mass of water is heated. The specific heat of water is 4,120 J/(Kg K). Calculate the electrical energy required to perform this process.

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Step 1

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mass of water,m=6000kg initial uniform temperature of themperature of water,T, = 20°C Final uniform temperature is T, =98°C specific heat of water 4120 j/(kgk heat loss through insulation=5x108j

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we know that Q=mc2dT Q=6000x 4120x (98-20) Q=6000 x 4120 x 78 Q=1.928 x 10°J According to question heat lost is 5x10°j

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Total electrical energy requried Q Q+o Q,(19.28+5)x 10 Q, = 24.28x 10 j

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