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how much heat must be removed from 456 gram of water at 25 degree Celsius to change it into ice at 10 degree Celsius in the specific heat of ice is 2090 joule per kilogram into came the latent heat of fusion of water is 33.5 X104 joule kg and specific heat of water is 4186 J/kg.K

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0.456 x 25 x 4186 + 33.5 x 104 x 0.456 +0.456 x 2090 x 10

= 0.456(25 x 4186+35.5 x104+2.09 x 104)

= 0.456(10.4 x 104 +33.5 x 104 +2.09 x 104)

= 0.456 x 45.99 x 104

=20.97 x 104 J

= 2.1 x 105 J

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