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Difficulty: EasyModes of Heat Transfer (Conduction, Convection, and Radiation)

A uniform metal rod of length 0.50 m0.50\text{ m} and cross-sectional area 2.0×10−3 m22.0 \times 10^{-3}\text{ m}^2 has a thermal conductivity of 400 W⋅m−1⋅K−1400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}. If a temperature difference of 50 K50\text{ K} is maintained between its ends, what is the rate of heat flow through the rod?

  1. 80 W80\text{ W}Answer
  2. B
    40 W40\text{ W}
  3. C
    20 W20\text{ W}
  4. D
    160 W160\text{ W}

Answer

The rate of heat flow through the rod is 80 W80\text{ W}.
The rate of heat transfer by conduction is calculated using Fourier's law, Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}. Substituting k=400 W⋅m−1⋅K−1k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, A=2.0×10−3 m2A = 2.0 \times 10^{-3}\text{ m}^2, ΔT=50 K\Delta T = 50\text{ K}, and d=0.50 md = 0.50\text{ m} yields 80 W80\text{ W}.

Step-by-Step Solution

1
Identify the given parameters for thermal conduction.
Thermal conductivity k=400 W⋅m−1⋅K−1k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, cross-sectional area A=2.0×10−3 m2A = 2.0 \times 10^{-3}\text{ m}^2, temperature difference ΔT=50 K\Delta T = 50\text{ K}, and length d=0.50 md = 0.50\text{ m}.
These parameters form the components of Fourier's law of thermal conduction.
2
Apply the conduction rate formula Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}.
\frac{Q}{t} = \frac{400 \times (2.0 \times 10^{-3}) \times 50}{0.50}
Heat transfer per unit time depends directly on conductivity, area, and temperature difference, and inversely on thickness/length.
3
Calculate the numerical value.
\frac{Q}{t} = \frac{40}{0.50} = 80\text{ W}
Dividing the product of the numerator terms (40 J/s⋅m40\text{ J/s}\cdot\text{m}) by length (0.50 m0.50\text{ m}) yields the rate of energy transfer.

Key Concept

Thermal Conduction Rate (Fourier's Law)
Estimated Time:45s
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Topics

Modes of Heat Transfer (Conduction, Convection, and Radiation)

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