Answer:
a)q= 2800 W/m²
b)To=59.4°C
Explanation:
Given that
L = 10 mm
K= 20 W/m·K
T=30°C
h= 100 W/m²K
Ti=58°C
a)
Heat flux q
q= h ΔT
q= 100 x (58 - 30 )
q= 2800 W/m²
b)
As we know that heat transfer by Fourier law given as
Q= K A ΔT/L
Lets take outer temperature is To
So by Fourier law
To= Ti + qL/K
Now by putting the values
To= Ti + qL/K
[tex]To= 58 + 2800 \times \dfrac{ 0.01}{20}[/tex]
To=59.4°C