Answer :
Answer:
Explanation:
The rate at which heat will be radiated is given by the expression
E = e Aσ ( T⁴ - T₀⁴ )
E is heat radiated , e is emissivity , A is area of surface , σ is stephan's constant T is temperature of the object and T₀ is temperature of the surrounding .
For all the objects given , e , σ T and T₀ are same so E will solely dependent on area of the surface
surface area of cube= 6 r² ,
surface area of sphere = 4 π r²
= 12.56 r²
hemisphere = 2 π r²
= 6.28 r²
12.56 r² >6.28 r² > 6 r²
heat radiated by sphere > heat radiated by hemisphere > heat radiated by cube .
Ranking the objects based on the net rate by which thermal radiation occurs
- Sphere ( 12.56 r² ) > Hemisphere ( 6.28 r² ) > Cube ( 6r² )
The net rate of thermal radiation is expressed as
E = e*A*∝ ( T₁⁴ - T₀⁴ ) ----- ( 1 )
where : e = emissivity , A = Area of object , ∝ = stephen constant, T₁ = object temperature , T₀ = temperature of surroundings.
From the question the values of e, T and ∝ are constant
Determine the net rate of thermal radiation ( E ) for each object
E = A --- ( 2 )
- For solid cube
E = A = 6r²
- For sphere
E= A = 4πr² = 12.56 r²
- For Hemisphere
E= A = 2πr² = 6.28 r²
Hence ranking the objects based on their thermal radiation,Sphere ( 12.56 r² ) > Hemisphere ( 6.28 r² ) > Cube ( 6r² )
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