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Zorluk: OrtaModes of Heat Transfer (Conduction, Convection, and Radiation)

A uniform metal cylinder of thermal conductivity 400 W⋅m−1⋅K−1400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1} has a cross-sectional area of 2.0×10−3 m22.0 \times 10^{-3}\text{ m}^2 and a length of 0.50 m0.50\text{ m}. Heat flows axially through the cylinder at a steady rate of 160 W160\text{ W}. Assuming no thermal losses through the lateral surface, what is the temperature difference between the two ends of the cylinder?

  1. 100 K100\text{ K}Cevap
  2. B
    200 K200\text{ K}
  3. C
    400 K400\text{ K}
  4. D
    50 K50\text{ K}

Cevap

The temperature difference between the two ends of the cylinder is 100 K100\text{ K}.
According to Fourier's law of heat conduction, the rate of heat transfer through a material is given by Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}. Rearranging the equation to solve for the temperature difference gives ΔT=(Q/t)⋅dkA\Delta T = \frac{(Q/t) \cdot d}{k A}. Substituting Q/t=160 WQ/t = 160\text{ W}, d=0.50 md = 0.50\text{ m}, k=400 W⋅m−1⋅K−1k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, and A=2.0×10−3 m2A = 2.0 \times 10^{-3}\text{ m}^2 yields ΔT=160×0.50400×2.0×10−3=800.8=100 K\Delta T = \frac{160 \times 0.50}{400 \times 2.0 \times 10^{-3}} = \frac{80}{0.8} = 100\text{ K}.

Adım Adım Çözüm

1
Identify the given thermal parameters and the rate of heat flow equation.
Rate of heat flow Qt=160 W\frac{Q}{t} = 160\text{ W}, 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, length d=0.50 md = 0.50\text{ m}. Formula: Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}.
Conduction through a solid body under steady-state conditions obeys Fourier's law of thermal conduction.
2
Rearrange the conduction formula to solve for the temperature difference ΔT\Delta T.
ΔT=(Qt)⋅dk⋅A\Delta T = \frac{\left(\frac{Q}{t}\right) \cdot d}{k \cdot A}.
Isolating ΔT\Delta T requires multiplying both sides by length dd and dividing by kAk A.
3
Substitute the values and compute ΔT\Delta T.
ΔT=160×0.50400×2.0×10−3=800.8=100 K\Delta T = \frac{160 \times 0.50}{400 \times 2.0 \times 10^{-3}} = \frac{80}{0.8} = 100\text{ K}.
Performing accurate arithmetic yields the temperature difference.

Anahtar Kavram

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

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