If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

Question

8 days ago

15

.5 cm thick and the thermal conductivity may be taken as 0.2 W/m K, what is the temperature difference across the material?

1 answer:

Maru [3.2K] · Answer 1 · 2026-04-02T08:38:09+03:00

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Consequently, the temperature difference across the material will be $\Delta T = 375 K$

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

$\Delta T$ is the temperature difference we need to determine

represents the thickness of the material $\Delta x=2.5 cm =0.025 m$

To isolate $\Delta T$ from equation (1), we obtain:

$\Delta T =\frac{Q \Delta x}{Ak}$

Initially, we convert 3KW to W, resulting in:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

With all variables accounted for, we can substitute and calculate:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Thus, the temperature difference across the material will be $\Delta T = 375 K$