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

Verified
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*}

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (1 more)Randall D. Knight
3,507 solutions

#### Applied Physics

11th EditionISBN: 9780134159386Dale Ewen, Erik Gundersen, Neill Schurter
3,468 solutions

#### Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260Mary L. Boas
3,355 solutions

#### Fundamentals of Physics

10th EditionISBN: 9781118230718 (2 more)David Halliday, Jearl Walker, Robert Resnick
8,971 solutions