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Chapter 11: Problem 23

If the temperature of the sun is increased from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64

Short Answer

Expert verified
The ratio of the radiant energy received on Earth to what it was previously is 64 times higher.

Step by step solution

01

Identify the relevant law

This is a problem involving the Stefan-Boltzmann Law, which states: The power emission is directly proportional to the fourth power of the temperature of the body and surface area. It is denoted as \(P = \sigma A T^4\), where \(P\) is the power emission, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the temperature.
02

Use the law to solve the problem

First, calculate the initial radiant energy received on Earth using the aforementioned law: \(P_i = \sigma (4 \pi R^2) T^4\). Next, calculate the radiant energy received after the changes, which will be \(P_f = \sigma (4 \pi (2R)^2) (2T)^4\), as the radius and temperature are doubled.
03

Calculate the ratio

The ratio between the two stages is then: \(\frac{P_f}{P_i}\), which simplifies to: \(\frac{64 T ^4\pi R^2}{T^4 \pi R^2} = 64\).
04

Conclusion

Therefore, if the temperature of the sun is doubled and its radius is doubled, the ratio of the radiant energy received on Earth to what it was previously would be 64 times.

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Most popular questions from this chapter

The rate of flow of thermal current depends on the nature of material (thermal conductivity), cross-sectional area \(A\), and temperature gradient. More the temperature difference, higher the thermal current flow. This fact identifies the thermal resistance offered by the material while conducting heat. One can find by equivalent resistance in heat flow using the same principles as for current. A body may transfer energy better by radiation. The nature of radiating surfaces play a role in the power radiated from it. On covering a surface by non-conducting/radiating media, the loss of the heat energy can be controlled. A rod of length \(l\) and conductivity \(k\) is placed between two reservoirs maintained at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\). At a distance \(\frac{l}{3}\) and \(\frac{l}{5}\) from \(100^{\circ} \mathrm{C}\) reservoir the rate of flow of thermal current are \(I_{1}\) and \(I_{2} .\) The temperatures at \(\frac{l}{5}\) and \(\frac{l}{3}\) are \(\theta_{1}\) and \(\theta_{2}\) then (A) \(I_{1}=I_{2}, 2 \theta_{1}=3 \theta_{2}\) (B) \(5 \theta_{1}-3 \theta_{2}=200\) (C) \(I_{1}=I_{2}, 5 \theta_{1}-3 \theta_{2}=200\) (D) \(I_{1} \neq I_{2}, \quad 5 \theta_{2}=3 \theta_{1}\)

If the temperature of the sun is increased from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64

When a hollow and a solid sphere of same material with same outer radius and identical surface finish are heated to the same temperature (A) In the beginning, both will emit equal amount of radiation per unit time (B) In the beginning, both will absorb equal amount of radiation per unit time (C) Both spheres will have same rate of fall of temperature \(\left(\frac{d T}{d t}\right)\) (D) Both spheres will have equal temperatures at any moment

A block body is at a temperature \(2880 \mathrm{~K}\). The energy radiation emitted by this object with wavelength between \(499 \mathrm{~nm}\) and \(500 \mathrm{~nm}\) is \(U_{1}\), between \(999 \mathrm{~mm}\) and \(1000 \mathrm{~nm}\) is \(U_{2}\), and between \(1499 \mathrm{~nm}\) and \(1500 \mathrm{~nm}\) is \(U_{3}\), then (Wien's constant \(b=2.88 \times 10^{6} \mathrm{~nm}-\mathrm{K}\) ) (A) \(U_{1}>U_{2}\) (B) \(U_{2}>U_{1}\) (C) \(U_{1}=0\) (D) \(U_{3}=0\)

A hot body is being cooled in air according to Newton's law of cooling, the rate of fall of temperature being \(k\) times the difference of its temperature with respect to that of surroundings. The time, after which the body will lose half the maximum heat it can lose, is (A) \(\frac{1}{k}\) (B) \(\frac{\ln 2}{k}\) (C) \(\frac{\ln 3}{k}\) (D) \(\frac{2}{k}\)
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