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A body at 80" "^(@)C cools down to 64" "^(@)C in 5 minutes, and in 10 minutes it cools down to 52" "^(@)C. What will be its temperature after 20 minutes ? What is the temperature of the environment ? |
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Answer» Solution :For the first 5 minute, `Deltat=5min`. `DeltaT=T_(2)-T_(1)=64-80=-16` Now USING `-(dQ)/(dt)=-k.[T-T_(s)]` `(16)/(5)=+k.[(80+64)/(2)-T_(s)]`. .. .(1) Here we have taken the average of initial and FINAL temperature as the temperature of the body. `T=(T_(1)+T_(2))/(2)` For the NEXT `Deltat=5` min., `(12)/(5)=k.((52+64)/(2)-T_(s))`. . . (2) Dividing equation (1) by equation (2) `(16)/(5)xx(5)/(12)=(72-T_(s))/(58-T_(s))` `:.4/3=(72-T_(s))/(58-T_(s))` `:.232-4T_(s)=216-3T_(s)` `:.T_(s)=16" "^(@)C` From equation (1) `(16)/(5)=k.(72-16)` `=56k.` `:.k.=(16)/(5xx56)=(2)/(32)` For the last `Deltat=10` minutes, `DeltaT=T-52`, where T = Final temperature `(52-T)/(10)=(2)/(35)((52+T)/(2)-16)(becauseTlt52)` `:.52-T=(4)/(7)((52+T-32)/(2))(because(20)/(35)=(4)/(7))` `:." "52-T=(2)/(7)(20+T)` `:.364-7T=40+2T` `:.324=9T` `:.T=(324)/(9)=36" "^(@)C` |
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