A sphere and a cube,both made of copper,have | vedclass

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

Question

(C) According to Stefan-Boltzmann law,the rate of loss of heat $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.Since both objects are black,mad

Vedclass · Accepted Answer

$(\frac{\pi}{6})^{\frac{1}{3}}$

Vedclass · Answer

(C) According to Stefan-Boltzmann law,the rate of loss of heat $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.Since both objects are black,made of the same material,and cooling in the same atmosphere at the same temperature,the rate of loss of heat is directly proportional to the surface area $A$.Let $V$ be the volume of both objects. For a sphere of radius $r$,$V = \frac{4}{3} \pi r^3$. For a cube of side $a$,$V = a^3$.Since $V$ is equal,$a^3 = \frac{4}{3} \pi r^3$,which implies $a = (\frac{4}{3} \pi r^3)^{1/3}$.The surface area of the sphere is $A_s = 4 \pi r^2$.The surface area of the cube is $A_c = 6 a^2 = 6 (\frac{4}{3} \pi r^3)^{2/3}$.The ratio of the rate of loss of heat is $\frac{A_s}{A_c} = \frac{4 \pi r^2}{6 (\frac{4}{3} \pi r^3)^{2/3}}$.Simplifying this,$\frac{A_s}{A_c} = \frac{4 \pi r^2}{6 (\frac{4}{3})^{2/3} \pi^{2/3} r^2} = \frac{4^{1/3} \pi^{1/3}}{6 (\frac{1}{3})^{2/3}} = \frac{4^{1/3} \pi^{1/3}}{6 \cdot 3^{-2/3}} = \frac{4^{1/3} \pi^{1/3} \cdot 3^{2/3}}{6} = \frac{(4 \cdot 9)^{1/3} \pi^{1/3}}{6} = \frac{(36)^{1/3} \pi^{1/3}}{6} = (\frac{36 \pi}{216})^{1/3} = (\frac{\pi}{6})^{1/3}$.

$A$ sphere and a cube,both made of copper,have equal volumes and are black. They are allowed to cool at the same temperature and in the same atmosphere. The ratio of their rate of loss of heat is:

Similar Questions

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.

Solar constant $(S)$ depends upon the temperature of the Sun $(T)$ as ........

If the radius of a star is $R$ and it acts as a black body,what would be the temperature of the star,in which the rate of energy production is $Q$? ($\sigma$ stands for Stefan's constant)

The total radiative power emitted by a spherical black body with radius $R$ and temperature $T$ is $P$. If the radius is doubled and the temperature is halved,then the radiative power will be

Two bodies $X$ and $Y$ at temperatures $T_1 \ K$ and $T_2 \ K$ respectively have the same dimensions. If their emissive powers are the same,the relation between their temperatures is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module