Homework Help: Cooling by thermal radiation

1. Mar 25, 2013

mstrakl

1. The problem statement, all variables and given/known data
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-Boltzman constant is 5,67×10^-8. There is no other radiation on the ball, from the surrounding area.
The ball's mass is 1kg.

2. Relevant equations

3. 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 boltzman 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?

2. Mar 25, 2013

tia89

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...

3. Mar 25, 2013

mstrakl

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

Attached Files:

• 20130325_182208.jpg
File size:
65 KB
Views:
190
4. Mar 25, 2013

tia89

I probably found your problem... when computing
$$\left(\frac{1}{T^3}\right)_{T/2}^{T}=\frac{1}{T^3/8}-\frac{1}{T^3}=\frac{8}{T^3}-\frac{1}{T^3}=\frac{1}{T^3}(8-1)$$
$$\frac{1}{T^3/8 -T^3}$$
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)

5. Mar 25, 2013

mstrakl

Oh of course.... I have found the problem now... Thanks :)