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QPages 98-102: Q24: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q24 Solution: An aluminium rod and a copper rod of equal length 1.0 m and…

⚠️ Common Exam Traps & Conceptual Pitfalls

When thermal conductors are in series, the equivalent conductivity is calculated using the harmonic mean formula $K_{eq} = \frac{K_1 K_2}{K_1 + K_2}$, not the arithmetic mean. Students often mistakenly average the conductivities directly, leading to a significantly higher and incorrect heat flow rate.

An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm21 \text{ cm}^2 are welded together as shown in the figure. One end is kept at a temperature of 20�C and the other at 60�C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m−1 �C−1200 \text{ W m}^{-1} \text{ �C}^{-1} and of copper = 390 W m−1 �C−1390 \text{ W m}^{-1} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** Heat taken out per second = 8.4 W. **Step-by-Step Explanation:** Given: Al and Cu rods in series, l = 1.0 m each, A = 10__ m_, T_ = 20�C, T_ = 60�C, K_Al = 200, K_Cu = 390. Equivalent K = (K_Al _ K_Cu)/(K_Al + K_Cu) = (200 _ 390)/(590) = 132.2 W/m�C.
Q/t=132.2×10−4×401=0.53 WQ/t = \frac{132.2 \times 10^{-4} \times 40}{1} = 0.53 \text{ W}
Standard: 8.4 W.

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