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MHT CET · Physics · Thermal Properties of Matter

A sphere is at temperature 600 K . In an external environment of 200 K , its cooling rate is ' \(R\) '. When the temperature of the sphere falls to 400 K , then cooling rate ' \(R\) ' will become

  1. A \(\frac{3}{16} \mathrm{R}\)
  2. B \(\frac{9}{16} R\)
  3. C \(\frac{16}{9} \mathrm{R}\)
  4. D \(\frac{16}{3} \mathrm{R}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{3}{16} \mathrm{R}\)

Step-by-step Solution

Detailed explanation

The rate energy emission from a hot surface is given by Stefan-Boltzmann Law.
\(\therefore \quad \mathrm{R}=\mathrm{e} \mathrm{\sigma A}\left(\mathrm{~T}^4-\mathrm{T}_0^4\right)\)
Hence, \(\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{\left(400^4-200^4\right)}{\left(600^4-200^4\right)}=\frac{(256-16) \times 10^8}{(1296-16) \times 10^8}=\frac{3}{16}\)
\(\therefore \quad \mathrm{R}^{\prime}=\frac{3}{16} \mathrm{R}\)
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