How can I calculate the cooling time of a steel ball using thermal radiation?

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Homework Help Overview

The discussion revolves around calculating the cooling time of a steel ball through thermal radiation. The original poster presents a problem involving a steel ball with specific properties and seeks to determine how long it takes to cool from 1700K to 850K using the Stefan-Boltzmann law.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up a differential equation based on the power of radiation and integrates it to find the cooling time. Some participants question the algebraic steps taken during the integration process and suggest checking for errors in the calculations.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's algebra and pointing out potential mistakes in the integration process. There is a recognition of differing results, with some participants suggesting that the correct approach should yield a time of approximately 313 seconds.

Contextual Notes

Participants are working under the assumption that the only cooling mechanism is thermal radiation, and there is a focus on the mathematical setup of the problem. The original poster has expressed confusion over the results, indicating a lack of clarity in the calculations.

mstrakl
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Homework Statement


Hi! I need some help by my phisycs homework at universty. Here is the problem:
We have a steel ball, with density of 7800kg/m^3, with speficic heat capacity 460J/kgK. Temperature of the ball is 1700K. How much time does the ball need to cool down to half of it's temperature (850K), if it cools only because of the thermal radiation. Stefan-Boltzmann constant is 5,67×10^-8. There is no other radiation on the ball, from the surrounding area.
The ball's mass is 1kg.

Homework Equations




The Attempt at a Solution


I suppose, that the power of radiation (P = j × A = (s.b. constant) × T^4 × A) is decreasing as the temperature drops, so i tried to set up my equaton with differentials:

dQ = Pdt
mc dT = A × (sb const.) × T^4 dt

i integrated the equation, and got this:
mcT^-4dT = A(sb const)dt
-1/3 mcT^-3= A(sb const) × t

integral is from Ta/2 to Ta

and i shrinked the expression and expressed time from the equation

t= 8mc/(Ta)^3×A×(sb. const)

; c= spec. heat capacity
sb const. = stefan Boltzmann constante = 5,67 × 10^-8
Ta = temp. at start = 1700K

the result i get is 51 secons. The solution should be 313 seconds. What am I doing wrong? Can you help me?
 
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mstrakl said:
mcT^-4dT = A(sb const)dt
-1/3 mcT^-3= A(sb const) × t

integral is from Ta/2 to Ta

and i shrinked the expression and expressed time from the equation

t= 8mc/(Ta)^3×A×(sb. const)

Check your algebra... when solving the integral you should have a 7 instead than 8... anyway (using of course 7) I have the right result...
 
I did the algebra again, I hope photo sharp enough
... i get the same result, I checked twice.

51,14 seconds.

The correct result should be 313 seconds :S
 

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I probably found your problem... when computing
$$ \left(\frac{1}{T^3}\right)_{T/2}^{T}=\frac{1}{T^3/8}-\frac{1}{T^3}=\frac{8}{T^3}-\frac{1}{T^3}=\frac{1}{T^3}(8-1) $$
instead you computed
$$ \frac{1}{T^3/8 -T^3} $$
which is clearly different because then you don't get a factor 1/8...
Using the right formula you should have 313 more or less (I did it before and it was OK) :smile:
 
Oh of course... I have found the problem now... Thanks :)
 

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