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An electric heater whose power is 54 W is immersed in `650cm^3` water in a calorimeter. In 3 min the water is heated by `3.4^@C`. What part of the energy of the heater passes out of the calorimeter in the form of radiant energy? |
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Answer» Energy suupplied by heater`=Pt`. `=54xx3xx60=9720J` Energy absorbed by water `=(650xx10^-3)xx4200xx3.4=9282J` Therefore, energy that passes out in the form of radiant heat `=9720-9282=438J` Percentage loss`=(438)/(9720)xx100=4.5%` |
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