Two spheres of same material have radii 1 m and 4 m and temperatures 4,000 K and 2,000 K respectively. The energy radiated per second by the first sphere is

According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is

P = σ A T⁴

Where,
– P is the power radiated,
– σ is the Stefan-Boltzmann constant,
– A is the surface area of the sphere
– T is the temperature of the sphere.

For a sphere, the surface area A is given by:

A = 4 π r²

Now, let’s calculate the energy radiated by both spheres:

For the first sphere:
– Radius r₁ = 1 m,
– Temperature T₁ = 4000 K.

Surface area:

A₁ = 4 π (1)² = 4 π m²

The power radiated:
P₁ = σ A₁ T₁⁴ = σ · 4 π · (4000)⁴

For the second sphere:
– Radius r₂ = 4 m,
– Temperature T₂ = 2000 K.

Surface area:

A₂ = 4 π (4)² = 64 π m²

The power radiated:

P₂ = σ A₂ T₂⁴ = σ · 64 π · (2000)⁴

Now, comparing P₁ and P₂:

– P₁ has a small surface area but a much greater temperature.
– P₂ has a larger surface area but a lower temperature.

However, the temperature factor dominates as the power varies as the fourth power of the temperature. So, 4000⁴ will be much larger than 2000⁴.

Hence, even though the surface area of the first sphere is smaller, the energy radiated per second by the first sphere will be greater than that by the second.

Conclusion:
The energy radiated per second by the first sphere is greater than that by the second.

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