Two spheres of radii 8 cm and 2 cm are cool | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy radiated by a body is given by Stefan-Boltzmann Law: $Q = A \varepsilon \sigma T^{4} t$.Since the time $t$,emissivity $\varepsilon$,and

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The energy radiated by a body is given by Stefan-Boltzmann Law: $Q = A \varepsilon \sigma T^{4} t$.Since the time $t$,emissivity $\varepsilon$,and Stefan-Boltzmann constant $\sigma$ are the same for both,the energy radiated $Q$ is proportional to $A T^{4}$.Since $A = 4 \pi r^{2}$,we have $Q \propto r^{2} T^{4}$.The temperatures must be in Kelvin: $T_{1} = 127 + 273 = 400 \ K$ and $T_{2} = 527 + 273 = 800 \ K$.The ratio of energy radiated is:$\frac{Q_{1}}{Q_{2}} = \left(\frac{r_{1}}{r_{2}}ight)^{2} \left(\frac{T_{1}}{T_{2}}ight)^{4}$$\frac{Q_{1}}{Q_{2}} = \left(\frac{8}{2}ight)^{2} \left(\frac{400}{800}ight)^{4}$$\frac{Q_{1}}{Q_{2}} = (4)^{2} 	imes \left(\frac{1}{2}ight)^{4}$$\frac{Q_{1}}{Q_{2}} = 16 	imes \frac{1}{16} = 1$.

Two spheres of radii $8 \ cm$ and $2 \ cm$ are cooling. Their temperatures are $127^{\circ} C$ and $527^{\circ} C$ respectively. Find the ratio of energy radiated by them in the same time.

Similar Questions

$A$ black body radiates energy at the rate of $E \ W/m^2$ at a high temperature $T \ K$. When the temperature is reduced to $\frac{T}{2} \ K$,the radiant energy will be

If the temperature of the body is increased by $10\%$,the percentage increase in the emitted radiation will be ....... $\%$

Find the radiant energy emitted per second in $Js^{-1}$ by a lamp filament at $2000 K$. The surface area is $5.0 \times 10^{-5} m^{2}$,the relative emissivity is $0.85$,and $\sigma = 5.7 \times 10^{-8} W m^{-2} K^{-4}$.

$A$ black body is at $727^\circ C$. It emits energy at a rate which is proportional to

$A$ black body radiates energy at the rate of $E \text{ W/m}^2$ at a high temperature $T \text{ K}$. When the temperature is reduced to $\left(\frac{T}{2}\right) \text{ K}$, the radiant energy is

Vedclass Test Series

Exam Paper Generator

Online Exam Module