Respuesta :
Answer: The value of the celsius temperature of the cube is 472.2°c.
Explanation:
The expression for the power radiated is as follows;
[tex]P=A\epsilon\sigma T^{4}[/tex]
Here, A is the area, [tex]\sigma[/tex] is the stefan's constant,[tex]\epsilon[/tex] is the emissivity and T is the temperature.
It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.
Then the expression for the radiated power for the cube and the sphere can be expressed as;
[tex]A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}[/tex]
Here, [tex]A_{1}[/tex] is the area of the sphere, [tex]A_{2}[/tex] is the area of the cube,[tex]T_{1}[/tex] is the temperature of the sphere and [tex]T_{2}[/tex] is the temperature of the cube.
The radiated powers and emissivity of the cube and the sphere are same.
[tex]A_{1}T_{1}^{4}=A_{2}T_{2}^{4}[/tex]
The area of the sphere is [tex]A_{1}=4\pi \times r^{2}[/tex].
Here, r is the radius of the sphere.
The area of the cube is [tex]A_{2}=6\times a^{2}[/tex].
Here, a is the edge of the cube.
Put [tex]A_{1}=4\pi \times r^{2}[/tex] and [tex]A_{2}=6\times a^{2}[/tex].
[tex]T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}[/tex] ....(1)
The masses and the densities of the sphere and the cube are same. Then the volumes are also same.
[tex]V_{1}=V_{2}[/tex]
Here,[tex]V_{1},V_{1}[/tex] are the volumes of the sphere and the cube.
[tex]\frac{4}{3}\pi r^{3}=a^{3}[/tex]
[tex]\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}[/tex]
Put this value in the equation (1).
[tex]T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}[/tex][tex]T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}[/tex]
Put T_{1}=500°c.
[tex]T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}[/tex]
[tex]T_{2}=472.2^{\circ}c[/tex]
Therefore, the value of the celsius temperature of the cube is 472.7°c.
