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A box with a total surface area of 1.20 m2 and a wall thickness of 4.00 cm is made of an insulatingmaterial. A 10.0-W electric heater inside the box maintains the inside temperature at 15.0°C abovethe outside temperature. Find the thermal conductivity k of the insulating materi

Question

A box with a total surface area of 1.20 m2 and a wall thickness of 4.00 cm is made of an insulatingmaterial. A 10.0-W electric heater inside the box maintains the inside temperature at 15.0°C abovethe outside temperature. Find the thermal conductivity k of the insulating materi

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Solution

The thermal conductivity (k) of a material can be calculated using the formula for the rate of heat transfer (Q/t), which is given by:

Q/t = kA(T1 - T2)/d

where:

  • Q/t is the rate of heat transfer (in watts),
  • k is the thermal conductivity (in W/m·K),
  • A is the surface area through which heat is being transferred (in m²),
  • T1 - T2 is the temperature difference across the material (in K), and
  • d is the thickness of the material (in m).

We can rearrange this formula to solve for k:

k = (Q/t)d/(A(T1 - T2))

Substituting the given values:

  • Q/t = 10.0 W,
  • A = 1.20 m²,
  • T1 - T2 = 15.0°C = 15.0 K (since a difference of 1°C is equal to a difference of 1 K), and
  • d = 4.00 cm = 0.04 m (since 1 cm = 0.01 m),

we get:

k = (10.0 W)0.04 m/(1.20 m²15.0 K) = 0.0222 W/m·K.

So, the thermal conductivity of the insulating material is approximately 0.0222 W/m·K.

This problem has been solved

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