Question

The thickness of a sheet of nickel is 0.4 cm. The
temperature difference between its two faces is 32°C.
It transmits heat through an area of 5 cm^2
at a rate
of 200 kcal per hour.
Calculate the coefficient of thermal
conductivity of nickel.​

Answer 1

Explanation Given The thickness of a sheet of nickel is 0.4 cm. The  temperature difference between its two faces is 32°C.  It transmits heat through an area of 5 cm^2  at a rate  of 200 k cal per hour.  Calculate the coefficient of thermal  conductivity of nickel.

Given thickness = 0.4 cm

Area = 5 cm^2

Q = 200 K cal

t = 60 x 60 = 3600 sec

We know that Q/t = kA (θ1 – θ2) / d

                       200 / 3600 = k x 5 x 10^-4 x 32 / 4 x 10^-3

                        0.055 = 160 x 10^-1 x k  

                         0.055  = 160 x 10^-1 k / 4

                            So k = 0.055 / 4

                    Or k = 0.0137 k cal s^-1 m^-1 deg C^-1