A lake is covered with a layer of ice under a steady sub-zero outside temperature. Let t₁ be the time taken for the thickness of the ice layer to increase from 1 cm to 2 cm , and t₂ be the time taken for the thickness to increase from 2 cm to 3 cm . The ratio t₁ : t₂ is

Step-by-step solution

The rate of heat conduction through an ice layer of thickness x is given by dQ dt = KA x . The heat released when a layer of thickness dx freezes is dQ = A L dx . Equating the two, we get A L dx dt = KA x dt = L KA x dx . Integrating both sides, the time taken for the thickness to increase from x_i to x_f is t = L 2KA (x_f^2 - x_i^2) . Thus, the time taken is proportional to the difference of the squares of the thicknesses. For the first interval: t₁ (2^2 - 1^2) = 4 - 1 = 3 . For the second interval: t₂ (3^2 - 2^2)