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^(@)C`. The diameter of the bubble at the bottom and at the surface are 3.6 mm and 4 mm respectively. Find the temperature of the lake at the bottom.

Explanation

Let `P_(1),V_(1),T_(1)andP_(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)+hrhog=(10^(5)+2.5xx10^(3)(10))N//m^(2)`
`P_(2)=10^(5)N//m^(2)" "V_(1)=4/3pir_(1)^(3),V_(2)=4/3pir_(2)^(3)`,
`T_(2)=273+40=313K`
From the equation, `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)`
`T_(1)=P_(1)/P_(2)xxV_(1)/V_(2)xxT_(2)`
`T_(1)=((10^(5)+0.25xx10^(5))xxcancel(4/3)cancel(pi)r_(1)^(3))/((10^(5))(cancel(4/3)cancel(pi)r_(2)^(3)))(313)`
= `1.242xx0.7288xx313=283.3K`
Temperature at the bottom of the lake
= `283.3-273=10.3^(@)C`