Examkin

Question

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

A cylindrical copper rod of thermal conductivity 380 W⋅m−1⋅K−1380\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, length 0.40 m0.40\text{ m}, and uniform radius 2.0 cm2.0\text{ cm} is thermally insulated along its curved surface. One flat end is maintained at a temperature of 100∘C100^\circ\text{C} by steam, while the opposite end is kept in an ice bath at 0∘C0^\circ\text{C}. Assuming steady-state heat conduction, what is the rate of heat flow through the rod? (Take π=3.14\pi = 3.14)

  1. 119.3 W119.3\text{ W}Answer
  2. B
    19.1 W19.1\text{ W}
  3. C
    1900.0 W1900.0\text{ W}
  4. D
    1193.2 W1193.2\text{ W}

Answer

The rate of heat flow through the copper rod is 119.3 W119.3\text{ W}.
According to Fourier's law of heat conduction, the rate of thermal energy transfer Qt\frac{Q}{t} through a material of thermal conductivity kk, cross-sectional area AA, and length dd across temperature difference ΔT\Delta T is given by Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}. Substituting k=380 W⋅m−1⋅K−1k = 380\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, A=π(0.02 m)2=1.256×10−3 m2A = \pi (0.02\text{ m})^2 = 1.256 \times 10^{-3}\text{ m}^2, ΔT=100 K\Delta T = 100\text{ K}, and d=0.40 md = 0.40\text{ m} yields 119.3 W119.3\text{ W}.

Step-by-Step Solution

1
Convert the radius from centimeters to meters and calculate the cross-sectional area of the rod.
r=2.0 cm=0.02 mr = 2.0\text{ cm} = 0.02\text{ m}. A=πr2=3.14×(0.02 m)2=1.256×10−3 m2A = \pi r^2 = 3.14 \times (0.02\text{ m})^2 = 1.256 \times 10^{-3}\text{ m}^2.
Fourier's law requires the area in square meters (m2m^2).
2
Determine the temperature difference across the rod.
ΔT=100∘C−0∘C=100 K\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100\text{ K}.
Heat transfer rate depends on the temperature gradient along the length.
3
Apply Fourier's law of thermal conduction: Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}.
Qt=380×1.256×10−3×1000.40=119.32 W≈119.3 W\frac{Q}{t} = \frac{380 \times 1.256 \times 10^{-3} \times 100}{0.40} = 119.32\text{ W} \approx 119.3\text{ W}.
Substituting the thermal conductivity kk, cross-sectional area AA, temperature difference ΔT\Delta T, and length dd gives the steady-state heat conduction rate.

Key Concept

Fourier's Law of Heat Conduction in solids: Qt=kA(Thot−Tcold)d\frac{Q}{t} = \frac{k A (T_{\text{hot}} - T_{\text{cold}})}{d}
Estimated Time:2m 0s
Rate this question

Topics

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

More questions from this topic