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Kshitij Kumar
Kshitij Kumar
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We have a solid and a hollow cubesphere made of same material, and of same dimensions. We have to compare the temperatures of the sphere after a long time of cooling.

So initially I can safely say that cooling by radiation will take place or: $$H=\epsilon A\sigma (\Delta T^4)$$

So initial rate of cooling will be same. However after inital conditions, the condcution will aslo take place inside the solid sphere, and here i start to have difficulty writing the equation for heat loss.

For the conduction, we have: $$H=\frac{KAdT}{dx}$$ Solving for a solid sphere case I get: $$H=KA\Delta T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

So we have conduction which is bringing in heat from beneath and heat leaving from the surface by radiation, so, how can I write net equation for the heat loss by the sphere and compare it quantitatively with hollow sphere? Answer is that solid sphere would stay warmer than the hollow one.

We have a solid and a hollow cube made of same material, and of same dimensions. We have to compare the temperatures of the sphere after a long time of cooling.

So initially I can safely say that cooling by radiation will take place or: $$H=\epsilon A\sigma (\Delta T^4)$$

So initial rate of cooling will be same. However after inital conditions, the condcution will aslo take place inside the solid sphere, and here i start to have difficulty writing the equation for heat loss.

For the conduction, we have: $$H=\frac{KAdT}{dx}$$ Solving for a solid sphere case I get: $$H=KA\Delta T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

So we have conduction which is bringing in heat from beneath and heat leaving from the surface by radiation, so, how can I write net equation for the heat loss by the sphere and compare it quantitatively with hollow sphere? Answer is that solid sphere would stay warmer than the hollow one.

We have a solid and a hollow sphere made of same material, and of same dimensions. We have to compare the temperatures of the sphere after a long time of cooling.

So initially I can safely say that cooling by radiation will take place or: $$H=\epsilon A\sigma (\Delta T^4)$$

So initial rate of cooling will be same. However after inital conditions, the condcution will aslo take place inside the solid sphere, and here i start to have difficulty writing the equation for heat loss.

For the conduction, we have: $$H=\frac{KAdT}{dx}$$ Solving for a solid sphere case I get: $$H=KA\Delta T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

So we have conduction which is bringing in heat from beneath and heat leaving from the surface by radiation, so, how can I write net equation for the heat loss by the sphere and compare it quantitatively with hollow sphere? Answer is that solid sphere would stay warmer than the hollow one.

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Kshitij Kumar
Kshitij Kumar
  • 2.1k
  • 8
  • 23

Cooling of a solid vs hollow sphere

We have a solid and a hollow cube made of same material, and of same dimensions. We have to compare the temperatures of the sphere after a long time of cooling.

So initially I can safely say that cooling by radiation will take place or: $$H=\epsilon A\sigma (\Delta T^4)$$

So initial rate of cooling will be same. However after inital conditions, the condcution will aslo take place inside the solid sphere, and here i start to have difficulty writing the equation for heat loss.

For the conduction, we have: $$H=\frac{KAdT}{dx}$$ Solving for a solid sphere case I get: $$H=KA\Delta T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

So we have conduction which is bringing in heat from beneath and heat leaving from the surface by radiation, so, how can I write net equation for the heat loss by the sphere and compare it quantitatively with hollow sphere? Answer is that solid sphere would stay warmer than the hollow one.