Question

A solid sphere has a temperature of 614 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins?

Answer 1

Answer:581.87 K

Step-by-step explanation:

Given

Sphere is melted to form a square

Let the radius of sphere be r and square has a side a

Therefore

`(4\pi)/(3)r^3=a^3`

Surface area of sphere
`A_s=4\pi r^2`

Surface area of cube
`A_c=6a^2`

Total emmisive remains same

Thus
`P=A\epsilon \sigma T^4`

`A_sT_s^4=A_cT_c^4`

`(T_c^4)/(T_s^4)=(A_s)/(A_c)`

`(T_c^4)/(T_s^4)=(1)/(2)* \left ( (4\pi)/(3)\right )^{(1)/(3)}`

`(T_c)/(T_s)=(1)/(2^(0.25))* \left ( (4\pi)/(3)\right )^{(1)/(12)}`

`T_c=T_s* (1)/(2^(0.25))* \left ( (4\pi)/(3)\right )^{(1)/(12)}`

`T_c=614* (1.12679)/(1.189)`

`T_c=581.87 K`