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A `10 kW` drilling machine is used to drill a bore in a small aluminium block of mass `8.0 kg`. How much is the rise in temperature of the block in `2.5` minutes, assuming `50%` of power is used up in heating the machine itself or lost to the surrounding? Specific heat of aluminium `= 0.91 J//g^(@)C`.A. `103^@C`B. `130^@C`C. `105^@C`D. `30^@C` |
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Answer» Correct Answer - A Total work done by drill machine in `2.5xx60s` `=(10xx10^3)(2.5xx60)=15xx10^5J` Energy lost `=50%` of `15xx10^5J=7.5xx10^5J` Energy taken by its surroundings, i.e., aluminium block `DeltaQ=mcDeltat=8xx10^3xx0.91xxDeltaTJ` Energy given `=`Energy taken `7.5xx10^5=8xx10^3xx0.91xxDeltaT` `impliesT=103^@C` |
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