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A body cools in 7 min from `60^@C` to `40^@C` What will be its temperature after the next 7 min? The temperature of surroundings is `10^@C`. |
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Answer» Rate of cooling `(triangleT)/(trianglet)=K(T-T_(0))` For cooling from `60^(@)C` to `40^(@)C` `implies(60-40)/(7)=K((60+40)/(2)-10)` `impliesK=(20)/(7xx40)=(1)/(14)` For cooling from `40^(@)C` to T `(40-T)/(7)=K((40+T)/(2)-10)implies(40-T)/(7)=(1)/(14)((40+T-20)/(2))` `implies160-4T=20+T` `implies140=5T` `impliesT=(140)/(5)=28^(@)C` `T=28^(@)C` |
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