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A copper ballscools from 62^(@) to 50^(@)Cin 10 mintuesin the next10 minutes . Calculate its temperatureat the endof further 10 minutes . |
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Answer» Solution :Theball cools from `theta_(1) = 62^(@) C ` to ` theta_(2) = 50^(@)C ` in `t_(1) = 10`minutes . (i) Average temperature during the interval ` theta = (theta_(1) + theta_(2))/(2) = 56^(@)C` Average rate of cooling `(d theta)/(dt) = (theta_(1) - theta_(2))/(t_(1)) = (62 - 50)/(10) = 1.2^(@) C min^(-1)` `(d theta)/(dt) prop (theta - theta_(0)) theta_(0)` = temperature of the surroudungs `therefore1.2 prop (56- theta_(0))` (ii)The ball cools from `theta_(2) = 50^(@)C , theta_(3) = 42^(@)C , t_(2)= 10` mintues . `theta^(1)= ( theta_(2) + theta_(3))/(2) = 46^(@)C , (d theta^(1))/(dt)= (theta_(2) - theta_(3))/(t_(2)) = (50 - 42)/(10) = 0.8^(@)C min^(-1)` `(d theta^(1))/(dt) prop(theta^(1) - theta_(0)) therefore0.8 (46-theta_(0)) "".........(2)` Divinding(1) by(2) we get`(1.2)/(0.8) = (56-theta_(0))/(46 - theta_(0))thereforetheta_(0) = 26^(@)C` The BALLS cools from `therefore _(3) = 42^(@) to therefore _(4) ` in `t_(3) = 10 ` minutes`therefore theta= (theta_(3) + theta_(4))/(2) = (42 + theta_(4))/(10), (d theta)/(dt) = (theta_(3) - theta_(4))/(t_(3)) = (42 - theta_(4))/(10)` Now `(42-theta_(4))/(10) prop ((42 + theta_(4))/(4)) - 26 ""......(3)` DIVIDING eq. (3) by (2) we have . `(((42-theta_(4))/(10))/0.8) = ((42+(theta_(4))/(2) - 26))/(46-26) "" therefore(42 - theta_(4))/( 8) = (42 + theta_(4) - 52)/(2 xx 20)` `5(42 - theta_(4)) = theta_(4)- 10 "" therefore 6 theta_(4) = 220 "" therefore _(4) = 36.7^(@)C` |
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