Question

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.

Answer 1

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

$Q=KA\dfrac{\Delta T}{t}$

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

$Q'=KA\dfrac{\Delta T}{t'}$

$Q'=KA\dfrac{\Delta T}{2t}$

$Q'=\dfrac{Q}{2}$

Therefore the new heat transfer become half of the original.