Question

A 10 cm thick slab of surface area 0.36 m 2 is placed on a hot surface at a constant temperature of 100°

c. a block of ice at 0 °c is placed on the upper surface of the slab. in half an hour, 2.4 kg of ice melts. find the thermal conductivity of the material of the slab. l of ice= 3.35 x 10 5 j/kg

Answer 1

According to  question

Area of Slab    A = 0.36 m^2m
Length              L = 10 Cm = 0.1 m
T1 = 0 deg C
T2 = 100 deg C

thus              delta T = T2 - T1
                                 = 100 deg C

Also       Q = 3.35 X 10^5

So    Thermal Conductivity     k = Q L / A delta T

                                              k = 3.35 X 10^5 * 0.1  /  0.36 * 100
                                             k = 935.55