Chapter 1.7, Problem 56P

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of $1m^2$, for a temperature difference of $20°C$.

To determine

The rate of heat conduction through a layer of still air.

Answer to Problem 56P

  $\frac{Q}{\Delta t}=-520\text{watts}$

Explanation of Solution

Given:

Thickness of air, $\Delta x=1\text{mm}=0.001\text{m}$

Area of the layer of air, $A=1{\text{m}}^2$

Temperature difference, $\Delta T=20°\text{C}$

Thermal conductivity of air, $k_t=0.026\text{J}{\text{.s}}^{\text{-1}}{\text{m}}^{\text{-1}}{\text{.K}}^{\text{-1}}$

Formula Used:

Rate of heat conduction:

  $\frac{Q}{\Delta t}=-k_{t}A\frac{\Delta T}{\Delta x}$

Where Q is the amount of heat, and $\Delta t$ is the time interval

Calculation:

The rate of heat conduction through a layer of still air is

  $\begin{array}{c}\frac{Q}{\Delta t}=-k_{t}A\frac{\Delta T}{\Delta x}\\ =-0.026\times 1\times \frac{20}{0.001}\\ =-520\text{watts}\end{array}$

Conclusion:

  $\frac{Q}{\Delta t}=-520\text{watts}$

The negative sign indicates that heat flows down the temperature gradient.