Question

A metal plate 4 mm thick has a temperature difference of
32°C between its faces. It transmits 200 kcal/h through an area of 5 cm². Calculate thermal conductivity of the material of the plate.​

Answer 1

Answer:

Kt = 58.5 W/m. K

Explanation:

The formula for thermal conductivity is

Kt =  delta Q / delta T x  L / A (T1 – T2)

Here Q is the heat energy , A the area of the surface, delta T is the overall time under consideration, L is the thickness of the metal plate.

As per given information we have;

delta Q / delta T = 200 kcal/h = 200000 x 4.184 / 3600

L / A (T1 – T2) = 4 / 5 (32)

Kt = 200000 x 4.184 / 3600 x 4 / 5 (32)

Solving the above, we get

Kt = 58.5 W/m. K

Answer 2

Answer:

Given :

Δ x = 4 mm = 4 × 10⁻³ m

Δ T = 32 C

Δ Q / Δ t = 200 k cal / hr

= > Δ Q / Δ t = ( 200 × 1000 × 4.2 ) / ( 60 × 60 ) J / sec = 233.33 J / sec

A = 5 cm² = 5 × 10⁻⁴ m²

We are asked to find Thermal conductivity i.e. K

We know :

K = ( Δ Q / Δ t ) / A ( Δ T / Δ x )

Putting values here we get :

K = ( 233.33 × 4 × 10⁻³ ) / ( 5 × 10⁻⁴ × 32 )

K =  ( 233.33 × 4 ) / ( 5 × 10⁻¹ × 32 )

K =   ( 233.33 × 4 ) / ( 5 × 3.2 )

K = 933.32 / 16

K = 58.33 W / m / C

Hence thermal conductivity of material of the plate is  58.33 W / m / C.