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:

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

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

Answer 2

The heat power for the baseball which is kept in the room temperature and moved to a region approximating T=0 K is 12.37 watts.

What is Stefan-Boltzman law?

According to the Stefan-Boltzman law, the amount of radiant heat power emitted from a surface, is directly proportional to the  fourth power of the absolute temperature of the surface.

It can be given as,

$P = 4\pi\varepsilon\sigmar^2T^4$

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).

The room temperature in kelvin will be 293.95 K.

Put the values in the above formula as,

$P = 4\pi(0.86)(5.67\times10^{-8})(0.052)^2(293.95)^4\\P=12.37\rm \; Watts$

Thus, the heat power for the baseball which is kept in the room temperature and moved to a region approximating T=0 K is 12.37 watts.

Learn more about the Stefan-Boltzman law here;

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