Answer:
Explanation:
Given
thermal conductivity of wood [tex] K_w=0.08 W/m^2-K[/tex]
thermal conductivity of insulation [tex]K_i=0.01 W/m^2-K[/tex]
thickness of wood [tex]t_2=3 cm[/tex]
thickness of insulation [tex]t_1=2.4 cm[/tex]
[tex]T_i=20^{\circ}C[/tex]
[tex]T_o=-13^{\circ}C[/tex]
we know heat Flow is given by
[tex]Q=kA\frac{dT}{dx}[/tex]
[tex]dT=[/tex] change in temperature
[tex]dx=[/tex] thickness
K=thermal conductivity
A=Area of cross-section
A is same
Suppose T is the temperature of Junction
as heat Flow is same thus
[tex]\frac{k_w(20-T)}{3}=\frac{k_i(T-(-13))}{2.4} [/tex]
[tex]\frac{0.08(20-T)}{3}=\frac{0.01(T+13)}{2.4}[/tex]
[tex]T=19.36 ^{\circ}C[/tex]
(b)Rate of heat flow
[tex]Q=\frac{k_w(T+13)}{3\times 10^{-2}}[/tex]
[tex]Q=\frac{0.08\times 32.36}{0.03}[/tex]
[tex]Q=86.303 W/m^2[/tex]