Question

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Answer 1

Question:- A closed cubical box is made of a perfectly insulating material and the only  way  for heat to enter or leave is through two solid cylindrical metal plugs, each of cross-sectional area 12cm² and length 8 cm fixed in the opposite walls of the box. The outer surface of  one plug is kept at 100°C while the outer surface of other plug is maintained at a temperature 4°C. The thermal conductivity of the material of the plug is 2 Wm⁻¹ °C⁻¹. A source of energy generating 13W is enclosed inside the box.  Find the  equilibrium temperature of the inner surface  of the box assuming that it is the same at all points on the inner surface.

Answer: 269 °C

Explanation:

Let the equilibrium temperature of the inner surface be Ф and K be Thermal conductivity of  the material of  the plug

K = 2 Wm⁻¹ °C⁻¹

Area of cylindrical plugs (A) = 12 cm²

A = 12 × 10⁻⁴ m²

Ф₁ = 100 °C

Ф₂ = 4 °C

Length of cylindrical plug from the surface of cubical box is  x which is distance travelled by heat.

x = 8 cm

Let that  the heat enters with rate of H₁ from the left cylindrical metal plug.

H₁ = KA ×(Ф₁-Ф)/x

Also  there is source of energy generating 13W  is enclosed inside the box.

Total heat enters in the  box = H₁ + 13 W

Let the rate at which  heat goes out be  H₂ from the right cylindrical plug.

H₂ = KA(Ф-Ф₂)/x

In the steady state, The heat comes inside will be equal to heat goes out.

Hence We have,

$H_1+13=H_2\\\;\\\frac{KA(\theta_1-\theta)}{x}+13=\frac{KA(\theta-\theta_2)}{x}\\\;\\\frac{KA\theta_1}{x}-\frac{KA\theta}{x}+13=\frac{KA\theta}{x}-\frac{KA\theta_2}{x}\\\;\\\frac{KA\theta_1}{x}+\frac{KA\theta_2}{x}+13=\frac{KA\theta}{x}+\frac{KA\theta}{x}\\\;\\\frac{KA(\theta_1+\theta_2)}{x}+13=\frac{2KA\theta}{x}\\\;\\\frac{2\times2\times12\times10^{-4}\times(100+4)}{0.08}+13=\frac{2\times12\times10^{-4}\times\theta}{0.08}\\\;\\3.12+13=0.06\times\theta\\\;\\\theta\times0.06=16.12$

$\theta=\frac{16.12}{0.06}\\\;\\\theta=268.66\\\;\\\theta\approx269\textdegree C$