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What amount of ice at `-14^(@)C` required to cool `200gg` of water from `25^(@)C` to 10CC? (Given `C_("ice")=0.5cal/g.^(@)C,L_(f)` for ice `=80cal//g`)A. `31g`B. `41g`C. `51g`D. `21g` |
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Answer» (A) Heat given by water in cooling from `25^(@)` to `10^(@)C` `Q_(1)=(cm DeltaT)_(w)=200xx1xx(25-10)=3000cal` Heat absorbed by `mg` of ice at `-14^(@)C` `Q_(2)=(mc DeltaT)_("ice")+mL_(f)+(mcDeltaT)_(w)` `=m[(0.5)[0-{-14}]+80+1(10-0)]` `=97mcal` `Q_(2)=Q_(1)` `97m=3000=30.93g` `m=31g` |
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