Answer:
T=120.04°C
Step-by-step explanation:
Given that
L= 7.5 cm
q = 105 W/m³
T∞=120°C
h=750 W/m²K
K=20 W/mK
Here given that one side of the wall is insulated that is why the maximum temperature will be at the insulated surface.
The total heat transfer from the wall
Q= q A L
Q= 150 x 0.075 A
Q=7.875 A W
A=Area of wall
Now the total thermal resistance R
![R=(0.00508)/(A)](https://img.qammunity.org/2020/formulas/engineering/college/6ub8b103n655exmv4jfb9efy8imazq5qqh.png)
We also know that
![Q=(\Delta T)/(R)](https://img.qammunity.org/2020/formulas/engineering/college/vdb2e71etvzg8gaqpmm7t88581j081vvze.png)
Temperature at insulated side = T
![7.875 =(T-120)/(0.00508)](https://img.qammunity.org/2020/formulas/engineering/college/dhujb1n41m7khpgjxvpigarlbyqagik8rn.png)
T=120.04°C