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If with steady-state heat flow established, you double the thickness of a wall built from solid uniform material, the rate of heat loss for a given temperature difference across the thickness will:

A) become 1/ 2 of its original value.
B) become one-half of its original value.
C) become four times its original value.
D) also double.
E) become one-fourth of its original value.

Respuesta :

Answer:

A) Become 1/ 2 of its original value.

B) Become one-half of its original value.

Explanation:

As we know that heat transfer trough the wall of thickness t given as

[tex]Q=KA\dfrac{\Delta T}{t}[/tex]

Where

K=Thermal conductivity of the wall

A= Cross sectional area of the wall

t=Thickness of the wall

ΔT=temperature difference across the wall surfaces.

When the thickness become double ,lets sat t' = 2 t

Then new heat transfer

[tex]Q'=KA\dfrac{\Delta T}{t'}[/tex]

[tex]Q'=KA\dfrac{\Delta T}{2t}[/tex]

[tex]Q'=\dfrac{Q}{2}[/tex]

Therefore the new heat transfer become half of the original.

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