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An electrically heated coil is immersed in a calorimeter containing 360 g of water at `10^@C`. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and coil is 40 g. The temperature of water after 10 min isA. `4.214^@C`B. `42.14^@C`C. `30^@C`D. none of these |
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Answer» Correct Answer - B Energy supplied by the heater to the system in 10 min `Q_1=Pxxt90 J//s xx10xx60s` `=54000J=(54000)/(4.2)cal=12857cal` Now if `theta` is the final temeprature of the system, energy absorbed by it to change its temperature from `10^@C` to `theta^@C` is `Q_2=(msDeltaT)_("water")+(msDeltaT)_("coil + calorimeter")` `=360xx1xx(theta-10)+40(theta-10)` `=400(theta-10)` According to problem, `Q_1=Q_2` So `12857=400(theta-10)` or `theta=42.14^@C` |
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