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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 uniform slab measuring 10 cm × 10 cm × 1 cm is placed between two...

A uniform slab measuring 10 cm×10 cm×1 cm10 \text{ cm} \times 10 \text{ cm} \times 1 \text{ cm} is placed between two heat reservoirs maintained at temperatures of 10^{}^{{}^{\circ}} \text{C} and 90^{}^{{}^{\circ}} \text{C}. The larger surface areas of the slab are in contact with the reservoirs. If the thermal conductivity of the material is 0.80 \text{ W m}^{-1} ^{}^{{}^{\circ}} \textC^{-1}, determine the amount of heat flowing through the slab per minute.

Options

  1. A64 J64 \text{ J}
  2. B384 J384 \text{ J}
  3. C4320 J4320 \text{ J}
  4. D3840 J3840 \text{ J}

Correct answer

D. 3840 J3840 \text{ J}

Step-by-step solution

Area of the larger surface, A=10 cm×10 cm=100 cm2=10−2 m2A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2 = 10^{-2} \text{ m}^2

Thickness of the slab, d=1 cm=10−2 md = 1 \text{ cm} = 10^{-2} \text{ m}

Temperature difference, \Delta T = 90^{}^{{}^{\circ}}\text{C} - 10^{}^{{}^{\circ}}\text{C} = 80^{}^{{}^{\circ}}\text{C}

Time, t=1 minute=60 st = 1 \text{ minute} = 60 \text{ s}

Thermal conductivity, K = 0.80 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

The amount of heat flowing through the slab is given by Fourier's law of heat conduction:

Q=KAΔTtdQ = \dfrac{K A \Delta T t}{d}

Substituting the given values:

Q=0.80×10−2×80×6010−2Q = \dfrac{0.80 \times 10^{-2} \times 80 \times 60}{10^{-2}}

Q=0.80×80×60Q = 0.80 \times 80 \times 60

Q=3840 JQ = 3840 \text{ J}
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