Answer:
21544 N
Explanation:
Let the atmospheric pressure be 101325 Pa
The side length of the cubic box:
[tex]V = d^3 = 0.051m^3[/tex]
[tex]d = \sqrt[3]{V} = \sqrt[3]{0.051} = 0.371 m[/tex]
The area of the cubic box:
[tex]A = d^2 = 0.371^2 = 0.1375 m^2[/tex]
20 C = 20 + 273 = 293 K
180 C = 180 + 273 = 453 K
As the volume of the air inside the closed cube is not changed, assume the ideal gas law we have
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
Where P1 = 101325 Pa and T1 = 293K are the original atmospheric pressure and temperature. P2 and T2 = 453 are the new pressure and temperature after the cube gets heat up
[tex]P_2 = P_1\frac{T_2}{T_1} = 101325\frac{453}{293} = 156656 Pa[/tex]
The net force on each side of the box it its pressure times side area
[tex]F = P_2A = 156656 * 0.1375 = 21544 N[/tex]
