Calculating Thermal Radiation: Power and Wavelength of a 47000K Object

[oldspice1212]

Consider a 47000K object that emits thermal radiation.

How much power does it emit per square meter?

What is its wavelength of peak intensity?What formula do I use to solve these questions?

 

[soulhunter]

http://en.wikipedia.org/wiki/Planck's_law

 

[oldspice1212]

useless

 

[rock.freak667]

How about trying Stefan-Boltzmann's law?

Were you taught any topics prior to being given this problem?

 

[Nickie]

To calculate the power emitted by a 47000K object, we can use the Stefan-Boltzmann Law which states that the power emitted per unit area (P) is equal to the Stefan-Boltzmann constant (σ) multiplied by the object's surface area (A) and its temperature raised to the fourth power (T^4). So, the formula would be P = σAT^4.

To calculate the wavelength of peak intensity (also known as the peak emission wavelength) of the object, we can use Wien's Law which states that the peak emission wavelength (λmax) is equal to the Wien's displacement constant (b) divided by the object's temperature (T). So, the formula would be λmax = b/T.

To solve these questions, we would need to know the surface area of the object and the values of the Stefan-Boltzmann constant (σ) and Wien's displacement constant (b). These values can be found in reference materials or online databases. Once we have these values, we can plug them into the respective formulas to calculate the power emitted per square meter and the wavelength of peak intensity for the 47000K object.