A metal rod of length (10 ~cm ) and area of cross-section (2.8 10⁻⁴ ~m ^2 ) is covered with a non-conducting substance. One end of it is maintained at (80^ C ), while the other end is put in ice at (0^ C ). It is found that (20 gm ) of ice melts in (5 ~min ). The thermal conductivity of the metal in Js" ( m ⁻¹ ~K ⁻¹ ) is (Latent heat of ice is (80 cal g ⁻¹ ).)

Step-by-step solution

Given, length of rod, (l=10 ~cm =0.1 ~m ) area of cross-section of rod, (A=2.8 10⁻⁴ ~m ^2 ) temperature at one end, (T₁=80^ C ) temperature at other end, (T₂=0^ C ) quantity of melted ice, ( m =20 gm ) time taken to melt ice, (t=5 ~min =300 sec ) and latent heat of ice, (s=80 cal g ⁻¹ ) Now, rate of the heat flow (= m s 4.184 t ) (= 20 80 4.184 300 =22.314 ~J / s ) Rate of heat flow in the rod, ( Q= k A T l ) ( aligned & 22.314= k (2.8 10⁻⁴ ) 80 0.1 & k=99.61 100 Js ⁻¹ ~m ⁻¹ ~K ⁻¹ aligned )