How much does the power radiated by a blackbody increase when its temperature (in \(\mathrm{K}\) ) is tripled?

Short Answer

Expert verified
The power radiated by a blackbody increases 81 times when its temperature is tripled.

Step by step solution

01

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law describes the power radiated by a blackbody at a certain temperature. The formula for the law is: \[P = \sigma A T^4\] where: - \(P\) is the power radiated by a blackbody, - \(\sigma\) is the Stefan-Boltzmann constant, equal to \(\approx 5.67 \times 10^{-8} \mathrm{W m^{-2}K^{-4}}\), - \(A\) is the surface area of the blackbody, - \(T\) is the temperature of the blackbody in Kelvins.
02

Calculate the initial power radiated by the blackbody

Let's consider the initial temperature of the blackbody as \(T_i\). The initial power radiated by the blackbody can be calculated as follows: \[P_i = \sigma A T_{i}^4\]
03

Calculate the final power radiated by the blackbody

Now the temperature of the blackbody is tripled, so the new temperature would be \(3T_i\). To calculate the final power radiated by the blackbody when the temperature is tripled, we can use the Stefan-Boltzmann Law again: \[P_f = \sigma A (3T_{i})^4\]
04

Find the increase in power radiated

To find the increase in power radiated by the blackbody, we need to compare the final power radiated (\(P_f\)) to the initial power radiated (\(P_i\)). We can do this by taking the ratio of \(P_f\) and \(P_i\): \[\frac{P_f}{P_i} = \frac{\sigma A (3T_{i})^4}{\sigma A T_{i}^4}\] Notice that \(\sigma A\) appears in both numerator and denominator, so it cancels out: \[\frac{P_f}{P_i} = \frac{(3T_{i})^4}{T_{i}^4}\] Now simplify the expression: \[\frac{P_f}{P_i} = 3^4 = 81\]
05

Express the result

Now we have found the increase in power radiated when the temperature is tripled: \[\frac{P_f}{P_i} = 81\] This means the power radiated by the blackbody, when its temperature is tripled, increases 81 times compared to its initial power.

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