A 2.0-cm-diameter metal sphere is glowing red, but a spectrum sho...

Question

A 2.0-cm-diameter metal sphere is glowing red, but a spectrum shows that its emission spectrum peaks at an infrared wavelength of 2.0 μm. How much power does the sphere radiate? Assume e=1 .

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, we have a spherical black body object that has a radius of 1.5 centimeters and an emission peak wavelength of 4.0 micrometers assuming the emissivity is one, calculate the power radiated. So that's our end goal. We're ultimately trying to figure out what the power radiated is and the power radiated will be our final answer. Awesome with that in mind, we're let's look at our multiple choice answers to see which answer might be our final answer. And let's also note that all of our multiple choice answers are in units of watts. So A is 33 B is 44 C is 40 D is 35. So first off, let us note that we are given the emission peak wavelength as so lambda P, which I'm gonna denote the peak wavelength as LAMBDA P is equal to 4.0 micrometers which is equal to 4000 nanometers. So now we must determine what the temperature is. In order to help us find our final answer, which we're trying to figure out what the power radiated is. So let us recall and use wine's law to find the temperature to find T. So let's recall that wine's law is lambda is equal to 2.90 multiplied by 10 to the power of six. And its units are nanometers multiplied by Kelvin divided by T. So we can rearrange wine's law to isolate and solve for tea. And when we do that, we will find that tea is equal to 2.90 multiplied by 10 to the power of six nanometers multiplied by Kelvin divided by lambda. So now let's plug in all of our known variables to solve for T which is our. So it's all for the temperature. I said the time but I meant temperature. So now isolating and solving for T our temperature, we will take 2.90 multiplied by 10 to the power of six nanometers multiplied by Kelvin divided by our wavelength which you let's use nanometers. So we could get those units to cancel out leaving us with Calvin, which is exactly what we want our temperature value to be in. So 4000 nanometers. So as you can see, our nanometer units will cancel out leaving us with units at Kelvin. So our final answer when we plug it into our calculator, it will be 725 Kelvin. Awesome. So now at this stage, we need to recall and use the Stefan Boltzmann Law equation to help us solve our final answer P which is the power radiated. So let us recall that Q which is the heat divided by the change in temperature is equal to the power radiated, which is equal to the emissivity multiplied by the Stefan Boltzmann Law or Stefen Boltzmann constant. So which is denoted as sigma, which is once again, the Stefan Boltzmann constant multiplied by the surface area of a black body multiplied by the temperature. Awesome. So with that in mind, let's quickly off to the side here in blue. Let's make a note here. Let us note that we are told that e the emissivity is equal to one. And let us also note that we can determine the surface area of the black body by using the area of a sphere because we're dealing with a spherical black body. So the area of a sphere states that four multiplied by pi multiplied by the radius squared. And let's note that the radius in this case is 1.5 centimeters which is equal to 0.015 m. Awesome. And of course, we know that T is equal to 725 Kelvin and we will, or we might know off the top of her head. But if you look it up, you'll know that the Stefan Boltzmann constant is equal to 5.67 multiplied by 10 to the power of negative eight and its units are watts her meters squared multiplied by Kelvin to the power of four. Awesome. So with that in mind, we can now plug in all of our known variables to solve for P which is our final answer the power radiated. So P equals one, since we know our emissivity is one multiplied by Stefan Boltzmann constant, just 5.67 multiplied by 10 to the power of negative eight. And its units are watts per meter squared, multiplied by Kelvin to the power of four, multiplied by four, multiplied by pi Since now, we're doing the area of a black body or surface area of a black body, I should say multiplied by 0.015 m squared, multiplied by 725. Kelvin to the power of four. So when we plug that into our calculator, we will find that P is equal to 44.29 watts, which is approximately equal to 44 watts. And that is our final answer. Hooray. We did it. Therefore, the spherical black body object will radiate approximately 44 watts of power. And this corresponds to multiple choice answer letter B. And once again, our final answer is 44 watts. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

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A 2.0-cm-diameter metal sphere is glowing red, but a spectrum shows that its emission spectrum peaks at an infrared wavelength of 2.0 μm. How much power does the sphere radiate? Assume e=1 .

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