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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

A room features a window of dimensions 1.0 m × 2.0 m equipped with...

A room features a window of dimensions 1.0 m×2.0 m1.0 \text{ m} \times 2.0 \text{ m} equipped with two glasspanes. Each glasspane has a thickness of 1 mm1 \text{ mm} and they are separated by an air gap of 1 mm1 \text{ mm}. Determine the rate of heat flow through the closed window if the temperature inside the room is 32^{}^{{}^{\circ}} C while the outside temperature is 40^{}^{{}^{\circ}} C. The thermal conductivity of the window glass is 1.0 \text{ J s}^{-1} \text{ m}^{-1} ^{}^{{}^{\circ}} C^{-1} and that of air is 0.025 \text{ J s}^{-1} \text{ m}^{-1} ^{}^{{}^{\circ}} C^{-1}.

Options

  1. A400 J s−1400 \text{ J s}^{-1}
  2. B190 J s−1190 \text{ J s}^{-1}
  3. C381 J s−1381 \text{ J s}^{-1}
  4. D8000 J s−18000 \text{ J s}^{-1}

Correct answer

C. 381 J s−1381 \text{ J s}^{-1}

Step-by-step solution

Area of the window, A=1.0×2.0=2.0 m2A = 1.0 \times 2.0 = 2.0 \text{ m}^2

Thickness of each glass pane, dg=1 mm=10−3 md_g = 1 \text{ mm} = 10^{-3} \text{ m}

Thickness of the air gap, da=1 mm=10−3 md_a = 1 \text{ mm} = 10^{-3} \text{ m}

Temperature difference, \Delta T = 40^{}^{{}^{\circ}} C - 32^{}^{{}^{\circ}} C = 8^{}^{{}^{\circ}} C

The system consists of two glass panes and one air gap connected in series. The equivalent thermal resistance is given by:

Req=Rg1+Ra+Rg2=dgKgA+daKaA+dgKgAR_{eq} = R_{g1} + R_a + R_{g2} = \dfrac{d_g}{K_g A} + \dfrac{d_a}{K_a A} + \dfrac{d_g}{K_g A}

Req=2dgKgA+daKaAR_{eq} = \dfrac{2d_g}{K_g A} + \dfrac{d_a}{K_a A}

Substituting the given values:

Req=2×10−31.0×2.0+10−30.025×2.0R_{eq} = \dfrac{2 \times 10^{-3}}{1.0 \times 2.0} + \dfrac{10^{-3}}{0.025 \times 2.0}

R_{eq} = 10^{-3} + \dfrac{10^{-3}}{0.05} = 10^{-3} + 20 \times 10^{-3} = 21 \times 10^{-3} ^{}^{{}^{\circ}} C \text{ W}^{-1}

The rate of heat flow is:

H=ΔTReq=821×10−3=800021≈381 J s−1H = \dfrac{\Delta T}{R_{eq}} = \dfrac{8}{21 \times 10^{-3}} = \dfrac{8000}{21} \approx 381 \text{ J s}^{-1}
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