## Related questions with answers

Question

Determine the thermal conductivity of a piece of building material $0.25 \mathrm{~in}$. thick that has an $R$ value of $1.6 \mathrm{~ft}^2{ }{\degree} \mathrm{F} / \mathrm{Btu} / \mathrm{h}$.

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 2$\textbf{Given}$

$L=0.25 \text{ in}= 0.02083 \text{ ft}$

$R=1.6 \text{ } \dfrac{ \text{ft}^2 \text { h }^\circ \text{F }}{\text{Btu}}$

The R value is defined as

$R= \dfrac{L}{K}$

where L is the thickness of the material and K is the thermal conductivity of the material

Reorganizing the formula we express K:

$K= \dfrac{L}{R}$

By including the given data we get:

$\begin{align*} K&=\dfrac{0.02083}{1.6}\\ &\boxed{K= 0.013 \text{ } \dfrac{\text{ Btu}}{\text{ ft} ^\circ \text{F h} } } \end{align*}$

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