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Chapter 14: Problem 85

One end of a cylindrical iron rod of length \(1.00 \mathrm{m}\) and of radius \(1.30 \mathrm{cm}\) is placed in the blacksmith's fire and reaches a temperature of \(327^{\circ} \mathrm{C} .\) If the other end of the rod is being held in your hand \(\left(37^{\circ} \mathrm{C}\right),\) what is the rate of heat flow along the rod? The thermal conductivity of yal iron varies with temperature, but an average between the two temperatures is $67.5 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$. (tutorial: conduction)

Short Answer

Expert verified
Answer: The rate of heat flow along the rod is approximately 10.46 W.

Step by step solution

01

Understand the formula for heat conduction

The formula for heat conduction in a cylindrical rod is given by: Q = -kA(ΔT)/L where: Q = rate of heat flow (W) k = thermal conductivity (W/(m*K)) A = cross-sectional area of the rod (m²) ΔT = temperature difference (K) L = length of the rod (m)
02

Calculate the cross-sectional area

Given the radius (r) of the cylindrical iron rod, we can calculate its cross-sectional area A using the formula: A = πr² The radius is given as 1.3 cm which is equal to 0.013 m (in SI units): A = π(0.013)² = 0.0005309 m²
03

Calculate the temperature difference

We are given the two temperatures, T₁ (fire) and T₂ (hand), so we can calculate the temperature difference as follows: ΔT = T₁ - T₂ T₁ = 327°C = 327K (since there is no difference when converting from Celsius to Kelvin) T₂ = 37°C = 37K ΔT = 327 - 37 = 290 K
04

Calculate the rate of heat flow

Now that we have all the values, we can simply plug them into the heat conduction formula from Step 1: Q = -kA(ΔT)/L k = 67.5 W/(m*K) A = 0.0005309 m² ΔT = 290 K L = 1.00 m Q = -67.5 × 0.0005309 × 290 / 1.00 = -10.46 W (approx.) Note that the negative sign indicates that the heat flow is from the hot end to the cold end. The rate of heat flow along the rod is approximately 10.46 W.

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