Answer:
100.76 °C
Explanation:
Let T₁ be the temperature of outer surface. Area of surface
A = 3.14 x ( 15/2 )²x 10⁻⁴ m² = 176.625 x 10⁻⁴ m²
For thermal conduction through a metallic surface the equation is
[tex]\frac{Q}{t} = \frac{KA(T_1-T_2)}{d}[/tex]
Q is heat passed which is 800 J here , t is time which is 1 s here, K is coefficient of thermal conductivity which is 237 wm⁻¹ A is area , T₁ and T₂ temperature of two sides , T₂ is 100 ° C .
Putting the values
800 = [tex]\frac{237\times176.625\times(T_1-100)}{ 0.4\times10^{-2}}[/tex]
T₁ = 100.76 °C
The temperature of the outer surface will be "100.76°C".
According to the question,
Thermal conductivity = 237 W/m⋅K
Diameter = 15 cm
Thickness = 0.4 cm
Heat passed = 800 W
Temperature of inner surface = 100°C
Let,
Temperature of outer surface be "[tex]T_1[/tex]".
Area of surface (A):
= [tex]3.14\times (\frac{15}{2} )^2\times 10^{-4}[/tex]
= [tex]176.625\times 10^4[/tex]
We know,
→ [tex]\frac{Q}{t} = \frac{KA(T_1-T_2)}{d}[/tex]
By substituting the values,
[tex]800= \frac{237\times 176.625\times (T_1-100)}{0.4\times 10^{-2}}[/tex]
[tex]T_1 = 100.76^{\circ} C[/tex]
Thus the above answer is appropriate.
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