HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q48 Solution: A cylindrical rod of length 50 cm and cross sectional area is…

QPages 98-102: Q48: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q48 Solution: A cylindrical rod of length 50 cm and cross sectional area is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently reverse the temperature difference in the Stefan-Boltzmann law, calculating $T_{end}^4 - T_{chamber}^4$ instead of $T_{chamber}^4 - T_{end}^4$, which yields a negative value and contradicts the positive heat flow from the hot chamber to the rod. Always ensure the radiation term represents heat absorbed by the blackened end from the hotter surroundings.

A cylindrical rod of length 50 cm and cross sectional area

1 \text{ cm}^2

is fitted between a large ice chamber at 0�C and an evacuated chamber maintained at 27�C as shown in the figure. Only small portions of the rod are inside the chamber and the rest is thermally insulated from the surrounding. The cross section going into the evacuated chamber is blackened so that it completely absorbs any radiation falling on it. The temperature of the blackened end is 17�C when steady state is reached. Stefan constant

\sigma = 6 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}

. Find the thermal conductivity of the material of the rod.

Step-by-Step Verified Solution

**Answer:** K = 3.5 W/m�C. **Step-BY-Step Explanation:** Given: l = 50 cm = 0.5 m, A = 1 cm_ = 10__ m_, T_ice = 0�C, T_chamber = 27�C, T_end = 17�C, _ = 6 _ 10__. Heat conducted = Heat radiated: KA(17_0)/0.5 = _A(T_end_ _ T_chamber_).

K \times 34 = 6 \times 10^{-8} \times (17^4 - 27^4) \times 10^8 \Rightarrow K = 3.5 \text{ W/m�C}

QPages 98-102: Q48: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q48 Solution: A cylindrical rod of length 50 cm and cross sectional area is…

Step-by-Step Verified Solution

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