A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 gram of ice per second. If the rod is replaced by another with half the length and double the radius of the first, the thermal conductivity of the material of second rod is ( 1 4 )^ th that of first, the rate at which ice melts in gram per second will be (Latent heat of ice = 80 cal/g)

Step-by-step solution

The rate of heat flow through a rod is given by H = KA T L . The rate of melting of ice is m = H L_f = KA T L L_f , where L_f is the latent heat of fusion. For the first rod, the rate of melting is: m₁ = K₁ r₁^2 T L₁ L_f = 0.1 g/s For the second rod, the given parameters are K₂ = K₁ 4 , r₂ = 2r₁ , and L₂ = L₁ 2 . The rate of melting for the second rod is: m₂ = K₂ r₂^2 T L₂ L_f Substituting the values of K₂ , r₂ , and L₂ : m₂ = ( K₁ 4 ) (2r₁)^2 T ( L₁ 2 ) L_f m₂ = ( K₁ 4 ) 4 r₁^2 T L₁ 2 L_f m₂ = K₁ r₁^2 T L₁ 2 L_f =