Calculating Heat Loss and Temperature Drop Using Stefan-Boltzmann's Law

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In summary, the problem involves estimating the rate of heat loss from two teapots, one ceramic and one shiny, each holding 0.50 L of tea at 99 degrees C. The rate of heat loss is calculated using the Stefan-Boltzmann Law and assuming the teapots are spheres. To estimate the temperature drop after 30 minutes, the formula H*t = E is used, where H is the rate of heat loss, t is the time period, and E is the energy lost by the object.
  • #1
munchy35
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Homework Statement



A ceramic teapot (epsilon(e) = 0.70) and a shiny one (epsilon(e) = 0.10) each hold 0.50 L of tea at 99 degrees C.
a) Estimate the rate of heat loss from each. Consider only radiation, and assume the surroundings are at 20 degrees C.

b.)Estimate the temperature drop after 30 min for each.

Homework Equations



Stefan-Boltzmann's Law

change in Q/change in T = epsilon*Stefan constant*Area (T^4 - T^4)

The Attempt at a Solution



My professor told us to assume that the tea pot is a sphere so we can solve for the area.

so V = 4/3pi R^3 = .05 m^3 R=.2285

A = 4pi R^2 = .6563



Ceramic = (.70)(5.67*10^-8)(.6563)((372^4)-(293^4))
Ceramic = 306.87 W

Shiny = (.1)*5.67*10^-8)(.6563)((372^4)-(293^4))
Shiny = 43.84

Then we're supposed to use two sig figs, which I did,and the program I used keeps telling me I am wrong.
 
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  • #2
okay so I realized .5 L = .005 m^3

but still I get 66 for ceramic and 9.4 for shiny.

it still says I am wrong
 
  • #3
nevermind. i got the right answer.

.5 L = .0005 m^3

gotta work on my conversions!
 
  • #4
now i have no idea how to approach part b once i have my answers for part a.
 
  • #5
hey I am working on the same problem too and i am stuck on part b and i don't know how to even approach this problem. did u figure it out yet?
 
  • #6
In the first part you've worked out the *rate* of heat loss, which is energy dissipated over time, right? So you have (with some notation abuse*):

[tex]H=\frac{E}{T}[/tex]

I may be wrong, but from there, knowing how long the heat was dissipating for (in SECONDS, of course), you should be able to get a change in energy and then temperature...

*Technically this should be a time derivative, as the rate of heat loss is dependent on temperature itself, but...
 
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  • #7
Yup, Oscur is right. You won't get an exact answer because H should be the time derivative and not E/T, but that's why the question says "estimate".
 
  • #8
im still confused. =( so what is the formula to figure out the temperature drop after 30mins? and are we suppose to use derivative to solve this problem? I am so sorry to bother you about this but I am really lost.

are we suppose to use this equation? ---> H = E/T
if so... what is H? E? T?

sooooooo lost.
 
  • #9
Ok, basically the rate at which something radiates heat depends on its temperature. This is evident from the form of the Stefan-Boltzmann Law, which has a term in T in it. As a result, to properly calculate the temperature after a given time, you'd need to integrate with respect to time.

As the question says "estimate", however, I'd assume you can just rearrange the equation I gave (in which H is the rate of heat loss, E is a total change in energy and T is a time period, sorry I should have defined my variables and possible the capital T was confusing matters. Call it "t" from now on.)

That way, you have: H*t = E where E is the energy lost by the object. Any more helpful?
 

1. What is Stefan-Boltzmann's Law?

Stefan-Boltzmann's Law is a physical law that describes the relationship between the temperature and the thermal radiation emitted by an object. It states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature.

2. How is Stefan-Boltzmann's Law used in science?

Stefan-Boltzmann's Law is used to calculate the total energy emitted by an object based on its temperature. It is commonly used in astrophysics to estimate the luminosity of stars and in engineering to design and optimize thermal systems.

3. What is a black body and why is it important in Stefan-Boltzmann's Law?

A black body is an idealized object that absorbs all radiation that falls on it and emits thermal radiation according to its temperature. It is important in Stefan-Boltzmann's Law because it serves as a theoretical standard against which the radiation properties of real objects can be compared.

4. How is Stefan-Boltzmann's Law related to the greenhouse effect?

The greenhouse effect is caused by the absorption and re-emission of thermal radiation by certain atmospheric gases, which traps heat and warms the Earth's surface. Stefan-Boltzmann's Law is used to calculate the amount of thermal radiation emitted by the Earth and its atmosphere, which plays a crucial role in the greenhouse effect.

5. What are some practical applications of Stefan-Boltzmann's Law?

Stefan-Boltzmann's Law has numerous applications in science and engineering, such as estimating the temperature of stars, designing thermal insulation materials, and calculating the efficiency of solar panels. It is also used in climate science to model the Earth's energy budget and predict the effects of climate change.

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