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A rod of length l (laterally thermally insulated) of the uniform cross-sectional area A consists of a material whose thermal conductivity varies with temperature as K = K 0 a + bT , where K 0 , a and b are constants. T 1 and T 2 < T 1 are the temperatures of two ends of the rod. Then, the rate of flow of heat across the rod is

Options

  1. AA K 0 b l a + b T 1 a + b T 2
  2. BAK 0 b l a + b T 2 a + b T 1
  3. CK 0 A b l &#8289; ln a + bT 1 a + bT 2
  4. DAK 0 b l ln a + b T 2 a + b T 1

Correct answer

C. K 0 A b l &#8289; ln a + bT 1 a + bT 2

Step-by-step solution

As, we know that, ⇒ dQ dt = - KA dT dx ⇒ dQ dt = - K 0 A a + bT dT dx On integrating both sides within the proper limits. dQ dt ∫ 0 l ⁡ dx = - K 0 A ∫ T 1 T 2 dT a + bT This gives, dQ dt = K 0 A b l ⁡ ln a + bT 1 a + bT 2

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