The top of an insulated cylindrical container is covered by a disc having emissivity $0.6$ and thickness $1\, cm$. The temperature is maintained by circulating oil as shown in the figure. If the temperature of the upper surface of the disc is $127^\circ C$ and the temperature of the surroundings is $27^\circ C$,then the radiation loss to the surroundings will be (Take $\sigma = \frac{17}{3} \times 10^{-8} \, W/m^2 K^4$)

  • A
    $595 \, J/m^2 \cdot s$
  • B
    $595 \, cal/m^2 \cdot s$
  • C
    $991.0 \, J/m^2 \cdot s$
  • D
    $440 \, J/m^2 \cdot s$

Explore More

Similar Questions

If the temperature of a black body increases from $7^{\circ}C$ to $287^{\circ}C$,then what is the ratio of the rate of energy emission?

Difficult
View Solution

Two spherical black bodies have radii $r_1$ and $r_2$. Their surface temperatures are $T_1$ and $T_2$. If they radiate the same power,then $\frac{r_2}{r_1}$ is:

The rate of radiation of a black body at $0^{\circ} C$ is $E \text{ J}s^{-1}$. The rate of radiation of the black body at $273^{\circ} C$ will be

Suppose the Sun is a spherical body of radius $r$ with a surface temperature of $t \, ^\circ C$. It radiates energy like a black body. The power received per unit area at a distance $R$ from the center of the Sun will be: ($\sigma$ is the Stefan-Boltzmann constant)

Difficult
View Solution

Two spheres $P$ and $Q$,of the same color,having radii $8 \; cm$ and $2 \; cm$ are maintained at temperatures $127^{\circ}C$ and $527^{\circ}C$ respectively. The ratio of energy radiated by $P$ and $Q$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo