Question

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. Calculate the Celsius temperature of the cube if it is to emit the same radiant powers the sphere​

Answer 1

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant,$\epsilon$ 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;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube,$T_{1}$  is the temperature of the sphere and $T_{2}$  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$.

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$.

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$.

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here,$V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$  

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$$T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.