Answer:
a. 14.5 W/
![m^(2)](https://img.qammunity.org/2021/formulas/chemistry/middle-school/jelmjz7kot3sm80pidpl7v5ncxu8balssm.png)
b. 58 W
Step-by-step explanation:
Thermal conductivity k = 0.029 W/m∙K
Temperature difference ΔT = 10 ℃
Thickness of sheet x = 20 mm = 0.02 m
area of sheet = 2m x 2m = 4
![m^(2)](https://img.qammunity.org/2021/formulas/chemistry/middle-school/jelmjz7kot3sm80pidpl7v5ncxu8balssm.png)
a. heat flux q = k x (ΔT/x)
q = 0.029 x (10/0.02)
q = 0.029 x 500 = 14.5 W/
![m^(2)](https://img.qammunity.org/2021/formulas/chemistry/middle-school/jelmjz7kot3sm80pidpl7v5ncxu8balssm.png)
b. Rate of heat transfer through the sheet Q = heat flux x area
Q = 14.5 x 4 = 58 W