Question

A meteorite 1.20 cm in diameter is so hot immediately after penetrating the atmosphere that it radiates 20.0 kW of power. (a) What is its temperature, if the surroundings are at $20.0^\circ\textrm{C}$ and it has an emissivity of 0.800? (b) What is unreasonable about this result? (c) Which premise or
assumption is responsible?

Final Answer

- $5.59 \times 10^3 \textrm{ K}$
- The melting point of granite is $1260^\circ\textrm{C}$. The meteorite temperature is much greater than this melting point. The meteorite would be a liquid.
- The 20 kW radiant power is unreasonably large.

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This is College Physics Answers with Shaun Dychko.
A meteorite with diameter 1.2 centimeters, which is 1.2 times 10 to the minus 2 meters, is radiating energy at a rate of 20 kilowatts. And so, I've written

*Q*over*T*is negative 20 times ten to the three watts. It's negative because this radiation is going to the environment. This formula that we have for*Q*over*T*is the formula for the rate at which an object absorbs energy from its environment. So, if it's in fact having a net emission to the environment, then this formula will give us a negative answer, because that's why we have negative 20 here. So, the temperature of the environment is 20 degrees Celsius, which in absolute terms is 293.15 Kelvin. And, the emissivity, we are told is 0.80. So, we have to figure out the temperature of the meteorite. So, we have this formula for the rate at which radiation is emitted from an object. It's Stefan Boltzmann's constant times emissivity times its surface area times the temperature of the environment to the power of four minus temperature of the object to the power of four. So, the surface area of a sphere, which we'll assume this meteorite is, is four pi times its radius squared. So, that's four pi times diameter over two, because radius is half the diameter, squared, and this makes pi*D squared*because this denominator two gets squared and then cancelled through this four here. So, we substitute pi*D squared*in place of area. And then, we distribute this into the brackets and we have*Q*over*T*is*Sigma**E*pi*D squared**T 2*to the power of four minus*Sigma**E*pi*D squared**T 1*to the power of four. And then, we bring this term to the left side by adding it to both sides. And, we bring this term to the right side by subtracting it from both sides. And so, we have a minus*Q*over*T*on the right and we have a positive*Sigma**E*pi*D squared**T 1*to the power of four on the left. Then, we divide both sides by the factors multiplying by*T 1*. And then, we take both sides to the power of one quarter to solve for*T 1*because*T 1*to the power of four to the power of 1 quarter will become*T1*. And, these factors cancelled in this term, so that's why you just see*T 2*to the power of four by itself. And, these factors have to show up underneath the*Q*over*T*term here. So, then we plug in numbers and we have 293.15 Kelvin is the temperature of the environment to the power of 4 minus negative 20 times 10 to the 3 watts over Stefan Boltzmann's constant times emissivity times pi times the radius squared, or this is the diameter, I should say, squared. That shouldn't be there. So it's just*Sigma**E*pi*D squared*. Okay. All of this is to the power of one quarter, which gives 5.59 times ten to the three Kelvin. So, that is very very hot and the melting point of granite, just to take an example of a rock that this meteorite might be made out of, is 1260 degrees Celsius, which in Kelvin is 1533. To write it in scientific notation, so we have times ten to the three. And, we can compare it with this times 10 to the 3, we can see that the meteorite is supposedly about five times as hot as the melting point of granite. And so, since that's so much hotter than the melting point of granite, we can assume that this meteorite will actually be a liquid. And, it's not going to be a meteorite at all because a meteorite is a term for a rock that actually hits the ground, otherwise it's just a meteor. And, this is what happens to most meteors that hit the Earth is that they melt and then burn up in the atmosphere. So, that's what's going to happen to this one, too. So, it's not reasonable to think that it radiates 20 kilowatts if there's an assumption that it will hit the ground. There we go.