Answer:
[tex] \Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K[/tex]
So then the difference of temperature across the material would be [tex] \Delta T = 375 K[/tex]
Explanation:
For this case we can use the Fourier Law of heat conduction given by the following equation:
[tex] Q = -kA \frac{\Delta T}{\Delta x}[/tex] (1)
Where k = thermal conductivity = 0.2 W/ mK
A= 1m^2 represent the cross sectional area
Q= 3KW represent the rate of heat transfer
[tex] \Delta T[/tex] is the temperature of difference that we want to find
[tex] \Delta x=2.5 cm =0.025 m[/tex] represent the thickness of the material
If we solve [tex]\Delta T[/tex] in absolute value from the equation (1) we got:
[tex] \Delta T =\frac{Q \Delta x}{Ak}[/tex]
First we convert 3KW to W and we got:
[tex] Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W[/tex]
And we have everything to replace and we got:
[tex] \Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K[/tex]
So then the difference of temperature across the material would be [tex] \Delta T = 375 K[/tex]
This question involves the concepts of Fourier's Law of Heat Conduction and thermal conuctivity.
The temperature difference across the material is "375°C".
According to Fourier's Law of Heat Conduction:
[tex]Q=KA\frac{\Delta T}{\Delta x}\\\\\Delta T=\frac{Q\Delta x}{KA}[/tex]
where,
ΔT = temperature difference = ?
Q = heat transfer rate = 3 KW = 3000 W
Δx = thickness = 2.5 cm = 0.025 m
A = cross-sectional area = 1 m²
K = thermal conductivity = 0.2 W/mK
Therefore,
[tex]\Delta T =\frac{(3000\ W)(0.025\ m)}{(0.2\ W/mK)(1\ m^2)}\\\\[/tex]
ΔT = 375°C
Learn more about Heat Conduction here:
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