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Question: Two copper spheres of radii \[6cm\] and \[12cm\] respectively are suspended in an evacuated enclosur...

Two copper spheres of radii 6cm6cm and 12cm12cm respectively are suspended in an evacuated enclosure. Each of them are at temperature 15C15^{\circ}C above the surrounding. The ratio of loss of heat is:

& A.2:1 \\\ & B.1:4 \\\ & C.1:8 \\\ & D.8:1 \\\ \end{aligned}$$
Explanation

Solution

To calculate the ratio of loss as heat, we can use the Stefan-Boltzmann law. Which gives the relationship between the radiation and the degree of the power. Since all the necessary quantities are given, we can substitute and calculate the ratio of heat emitted.

Formula used:
ET4E\propto T^{4} and E=σAΔT=4πr2σΔTE=\sigma A{\Delta T}=4\pi r^{2} \sigma \Delta T

Complete answer:
Here, we have two spheres of the same material i.e. copper. Given that their radii are r1=6cmr_{1}=6cm and the other is r2=12cmr_{2}=12cm.

We know that radiation is the degree of how much power is emitter, reflected or transmitted by anybody.
Similarly we know that, according to Stefan-Boltzmann law, which states that, the total radiation of heat emitted from a surface is proportional to the fourth power of its absolute temperature.
ET4E\propto T^{4}
E=σT4E=\sigma T^{4}, where σ\sigma is the constant of proportionality called the Stefan-Boltzmann constant also, σ=5.670×108W/mK4\sigma=5.670\times 10^{-8} W/mK^{4}
Also given as E=σAΔT=4πr2σΔTE=\sigma A{\Delta T}=4\pi r^{2} \sigma \Delta T
Then the ratio is given as E1E2=4πr12σΔT4πr22σΔT\dfrac{E_{1}}{E_{2}}=\dfrac{4\pi r^{2}_{1} \sigma \Delta T}{4\pi r^{2}_{2} \sigma \Delta T}
Since the ΔT\Delta T for both the spheres are 15C15^{\circ} C, we get, the ratio of heat loss as E1E2=r12r22\dfrac{E_{1}}{E_{2}}=\dfrac{r^{2}_{1}}{r^{2}_{2}}
Substituting the values for r1=6cmr_{1}=6cm and r2=12cmr_{2}=12cm, we get
E1E2=62122=(12)2=14\dfrac{E_{1}}{E_{2}}=\dfrac{6^{2}}{12^{2}}=\left(\dfrac{1}{2}\right)^{2}=\dfrac{1}{4}

So, the correct answer is “Option B”.

Additional Information:
The Stefan-Boltzmann law talks about the heat emitted due to a black body in terms of temperature of the body. According to it, the total energy radiated per unit area of the black body across the wavelengths per unit time is proportional to temperature raised to the fourth power. It uses the basics of thermodynamics and Planck’s law.
This law is used to determine the temperature of the sun’s surface, radiations emitted by the stars and effective temperature of the earth.

Note:
The Stefan-Boltzmann law gives the radiation of heat produced due to a mass of an object. Also note that, since both the bodies are at same temperature, EAE\propto A, where AA is the area of the body. This law is used to determine the temperature of the sun’s surface, radiation emitted by the stars and effective temperature of the earth.