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QPages 98-102: Q10: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q10 Solution: The left end of a copper rod length = 20 cm, area of cross…

⚠️ Common Exam Traps & Conceptual Pitfalls

When using the formula $Q= rac{kADelta T}{L}$, all quantities must be in SI units. If you plug the area in $ ext{cm}^2$ or the length in $ ext{cm}$ directly, the heat current will be off by a factor of $10^4$. Convert $A=0.20, ext{cm}^2$ to $2 imes10^{-5}, ext{m}^2$ and $L=20, ext{cm}$ to $0.20, ext{m}$ before computing $Q$.

The left end of a copper rod length = 20 cm, area of cross section = 0.20 cm20.20 \text{ cm}^2) is maintained at 20�C and the right end is maintained at 80�C. Neglecting any loss of heat through radiation, find a) the temperature at a point 11 cm from the left end and b) the heat current through the rod. Thermal conductivity of copper = 385 W m−1 �C−1385 \text{ W m}^{-1} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** (a) T = 53�C (b) Heat current = 2.31 W. **Step-by-Step Explanation:** Given: length = 20 cm, A = 0.20 cm_, T_ = 20�C, T_ = 80�C, K = 385 W m__�C__. (a) At 11 cm from cold end: T = 20 + (80_20) _ (11/20) = 53�C. (b) Heat current = (385 _ 0.20 _ 10__ _ 60)/(0.20) = 2.31 W [citation:14].

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