Question

A baseball of radius r = 5.2 cm is at room temperature T = 20.8 C. The baseball has emissivity of ε = 0.86 and the Stefan-Boltzman constant is σ = 5.67 × 10-8 J/(s⋅m2⋅K4). Say this ball is now moved to a region approximating T=0 K (space, for instance). P = 4πεσr2T4. What is the power in W? Numeric : A numeric value is expected and not an expression. P =

Answer 1

Answer:

$P = 12.37 \frac{J}{s} = 12.37 Watts$

Explanation:

Previous concepts

The Thermal radiation is one of "3 mechanisms who allows to bodies exchange energy".

The thermal radiation formula is given by:

$\frac{P}{A} = \epsilon \sigma T^4$

Where $\sigma = 5.67 x10^{-8} \frac{J}{sm^2 K^4}$

If we solve for P we got:

$P = A \epsilon \sigma T^4$

Since we have a baseball ball considered as a sphere the superficial area is given by:

$A = 4\pi r^2$

Solution to the problem

And if we replace this into our equation of P we got:

$P = (4\pi r^2) \epsilon \sigma T^4$

And we can reorder this like that:

$P = 4 \epsilon \pi \sigma r^2 T^4$

We can convert the radius to meters and we got:

$r= 5.2 cm*\frac{1m}{100 cm}=0.052 m$

Now we can convert the temperature to Kelvin and we got:

$T = 20.8 +273.15 = 293.95 K$

$\epsilon = 0.86$ the emissivity given

And now we can replace into the formula for P and we got:

$P = 4*0.86*\pi *(5.67x10^{-8} \frac{J}{s m^2 K^4}) (0.052m)^2 (293.95 K)^4$

$P = 12.37 \frac{J}{s} = 12.37 Watts$