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A metal plate 4 mm thick has a temperature difference of 32°C between its faces. It transmits 200 kcal/h through an area of 5 cm². Calculate thermal conductivity of the material of the plate. |
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Answer» Given : Δ x = 4 mm = 4 × 10⁻³ mΔ T = 32 CΔ Q / Δ t = 200 k cal / hr= > Δ Q / Δ t = ( 200 × 1000 × 4.2 ) / ( 60 × 60 ) J / sec = 233.33 J / secA = 5 cm² = 5 × 10⁻⁴ m²We are ASKED to find Thermal conductivity i.e. KWe know : K = ( Δ Q / Δ t ) / A ( Δ T / Δ x ) Putting values here we get : K = ( 233.33 × 4 × 10⁻³ ) / ( 5 × 10⁻⁴ × 32 ) K = ( 233.33 × 4 ) / ( 5 × 10⁻¹ × 32 ) K = ( 233.33 × 4 ) / ( 5 × 3.2 ) K = 933.32 / 16K = 58.33 W / m / CHence thermal conductivity of material of the plate is 58.33 W / m / C. |
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