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An air bubble comes from the bottom to the surface of a lake of depth 2.5 m. The surface temperature of the lake is 40^(0)C. The diameter of the bubble at the bottom a,id at the surface are 3. 6 mm and 4 ,nni respectively. Find the temperature of the lake at tlie bottom

Answer»

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Solution :let `P_(1), V_(1) , T_(1) and P_(2) , V_(2) , T_(2)` are the parameters of the air BUBBLE at the bottom and SURFACE of the lake respectively,
`P_(1) = P_(0) +h rho g = (10^(5) + 2.5 xx 10^(3) (10)) N//m^(2)`
`P_(2) = 10^(5) N//m^(2) "" V_(1) = (4)/(3) pi r_(1)^(3), V_(2)= (4)/(3) pi r_(2)^(3),`
`T_(2) = 273 + 40 = 314 `K
From the equation , `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
`T_(1) = (P_(1))/(P_(2)) xx (V_(1))/(V_(2)) xx T_(2)`
`T_(1) = ((10^(5) + 0.25 xx 10^(5)) xx cancel((4)/(3))cancel(pi)r_(1)^(3))/((10^(5)) (cancel((4)/(3))cancel(pi)r_(2)^(3))) (313)`
`1.242 xx 0.7288 xx 313 = 283.3 K `
Temperature at the bottom of the lake
= 283.3- 273 = `10.3^(0)` C


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