Temperature of water at the surface of lake is – 20°C. Then temperature of water just below the lower surface of ice layer is

Explanation

The situation is show in figure. Let the temperature inside the box be (theta). The rate at which heat enters the box through the left plug is
`(DeltaQ_(1))/(Deltat)=(KA(theta_(1)-(theta)))/(x)` . The rate of heat generation in the box `=13W` . The rate at which heat fkows out of the box through the right pluyg is `(DeltaQ_(2))/(Deltat)=(KA(theta-theta_(2)))/(x)` . In the steady state
`(DeltaQ_(1))/(Deltat)+13W=(DeltaQ_(2))/(Deltat)` . or, `(KA)/(x)(theta_(1)-theta)+13W=(KA)/(x)(theta-theta_(2))` . or, `2(KA)/(x)theta=(KA)/(x)(theta_(1)+theta_(2))+13W` . or, `theta=theta_(1)+theta_(2)/(2)+((13W)xx0.08m)/(2xx(2.0Wm^(-1)`^(@)C^(-1)(12xx10^(-4)m^(2))` . `=52^(@)C+216.67^(@)C=269^(@)C` .