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A body cools from 62^(@)C to 50^(@)C in 10 minutes and to 42^(@)C in the next 10 minutes. Find the temperature of surroundings. |
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Answer» Solution :For the FIRST ten minutes `(dT)/(dt)=-((62^(@)-50^(@))/10)=-1/2^(@)C//"min"` and `DeltaT=((62^(@)+50^(@))/2)-T_(0)(56-T_(0))^(@)C` `implies-1.2^(@)C//"min"=-KA(56-T_(0))^(@)C`………1 Similarly for the next ten minutes `(dT)/(dt)=[(42^(@)-50^(@))/2]=-0.8^(@)C`/min and `DeltaT=((42^(@)+50^(@))/2)-T_(0)=(45-T_(0))^(@)C` `implies-0.8^(@)` C/min=`-KA(45-T_(0))^(@)C`........2 DIVIDING 1 and 2 `3/2=(56^(@)-T_(0))/(46^(@)-T_(0))` `impliesT_(0)=26^(@)C` |
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