Question 734: Thermodynamics & KTG

Question

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

Accepted Answer

The rate of energy emission from a hot surface is described by the Stefan-Boltzmann Law, which can be expressed as:
$ R = e \sigma A (T^4 - T_0^4) $
In this formula:
$ R $ is the cooling rate,
$ e $ is the emissivity of the surface,
$ \sigma $ is the Stefan-Boltzmann constant,
$ A $ is the surface area of the sphere,
$ T $ is the temperature of the sphere,
$ T_0 $ is the temperature of the surrounding environment.
To determine the new cooling rate $ R' $ when the sphere's temperature decreases from 600 K to 400 K, we use the ratio:
$ \frac{R'}{R} = \frac{(400^4 - 200^4)}{(600^4 - 200^4)} $
Calculating further:
$ \frac{R'}{R} = \frac{(256 - 16) 	imes 10^8}{(1296 - 16) 	imes 10^8} = \frac{240 	imes 10^8}{1280 	imes 10^8} = \frac{3}{16} $
Thus, the new cooling rate is:
$ R' = \frac{3}{16} R $

Answer

$\frac{3}{16} \mathrm{R}$

Answer

$\frac{9}{16} R$

Answer

$\frac{16}{9} R$

Answer

$\frac{16}{3} R$

Thermodynamics & KTG

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

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