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

A glass window pane has an area of 1.50 m21.50\text{ m}^2 and a thickness of 4.0 mm4.0\text{ mm}. The inner and outer surface temperatures of the glass are maintained at 25∘C25^\circ\text{C} and 5∘C5^\circ\text{C} respectively. If the thermal conductivity of glass is 0.80 W⋅m−1⋅K−10.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, what is the rate of heat transfer through the glass pane by conduction?

  1. 6000 W6000\text{ W}Cevap
  2. B
    4000 W4000\text{ W}
  3. C
    87900 W87900\text{ W}
  4. D
    6.0 W6.0\text{ W}

Cevap

The rate of heat transfer through the glass pane is 6000 W6000\text{ W}.
According to Fourier's law of thermal conduction, the rate of heat flow Qt\frac{Q}{t} is given by Qt=kA(T1−T2)d\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d}. Substituting k=0.80 W⋅m−1⋅K−1k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, A=1.50 m2A = 1.50\text{ m}^2, T1−T2=20 KT_1 - T_2 = 20\text{ K}, and d=0.004 md = 0.004\text{ m} gives Qt=0.80×1.50×200.004=6000 W\frac{Q}{t} = \frac{0.80 \times 1.50 \times 20}{0.004} = 6000\text{ W}.

Adım Adım Çözüm

1
Identify the given values and convert all quantities to SI units.
Area A=1.50 m2A = 1.50\text{ m}^2, thermal conductivity k=0.80 W⋅m−1⋅K−1k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, thickness d=4.0 mm=4.0×10−3 md = 4.0\text{ mm} = 4.0 \times 10^{-3}\text{ m}, and temperature difference ΔT=25∘C−5∘C=20 K\Delta T = 25^\circ\text{C} - 5^\circ\text{C} = 20\text{ K}.
Fourier's law requires all linear dimensions to be in meters and temperatures in kelvins/degrees Celsius consistently.
2
Apply Fourier's law of thermal conduction equation.
Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}
The rate of heat conduction is directly proportional to thermal conductivity, surface area, and temperature difference, and inversely proportional to material thickness.
3
Substitute the values into the formula and solve.
Qt=0.80×1.50×204.0×10−3=240.004=6000 W\frac{Q}{t} = \frac{0.80 \times 1.50 \times 20}{4.0 \times 10^{-3}} = \frac{24}{0.004} = 6000\text{ W}
Calculates the total heat energy conducted per second across the window pane.

Anahtar Kavram

Thermal Conduction Rate Equation
Tahmini Süre:1m 30s
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Konular

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

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