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QPages 98-102: Q5: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q5 Solution: One end of a steel rod ) of length 1.0 m is kept in ice at 0�C…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget to convert the area from cm² to m², using $0.04\,\text{cm}^2$ as $0.04\,\text{m}^2$ and thus over‑estimating the heat flow by a factor of $10^4$. Always change $A=0.04\,\text{cm}^2$ to $4\times10^{-6}\,\text{m}^2$ before applying $\dot Q=KA\Delta T/L$.

One end of a steel rod K=46 J s−1 m−1 �C−1K = 46 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1}) of length 1.0 m is kept in ice at 0�C and the other end is kept in boiling water at 100�C. The area of cross section of the rod is 0.04 cm20.04 \text{ cm}^2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** Mass of ice melted per second = 2.74 _ 10__ kg/s. **Step-by-Step Explanation:** Given: K = 46 J s__ m__�C__, L = 1.0 m, A = 0.04 cm_ = 4 _ 10__ m_, _T = 100�C, L_fusion = 3.36 _ 10_ J/kg.
ΔQΔt=46×4×10−6×1001=0.0184 J/s\frac{\Delta Q}{\Delta t} = \frac{46 \times 4 \times 10^{-6} \times 100}{1} = 0.0184 \text{ J/s}
Mass melted per second = 0.0184/3.36 _ 10_ = 5.48 _ 10__ kg/s. Standard: 2.74 _ 10__ kg/s.

All Questions in this Chapter

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