Spheres P and Q are uniformly constructed fro | vedclass

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

Question

(C) Let the radius of sphere $P$ be $r$ and the radius of sphere $Q$ be $3r$.The rate of heat loss by radiation is given by Stefan-Boltzmann Law: $\fr

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let the radius of sphere $P$ be $r$ and the radius of sphere $Q$ be $3r$.The rate of heat loss by radiation is given by Stefan-Boltzmann Law: $\frac{dQ}{dt} = \sigma e A T^4$.Also,the rate of heat loss is related to the rate of fall of temperature by the equation: $\frac{dQ}{dt} = -ms \frac{dT}{dt}$,where $m$ is mass,$s$ is specific heat capacity,and $\frac{dT}{dt}$ is the rate of fall of temperature.Equating the two: $ms \frac{dT}{dt} = \sigma e A T^4$.Since $m = \rho V = \rho (\frac{4}{3} \pi r^3)$,we have $\rho (\frac{4}{3} \pi r^3) s \frac{dT}{dt} = \sigma e (4 \pi r^2) T^4$.Thus,the rate of fall of temperature is $\frac{dT}{dt} = \frac{\sigma e (4 \pi r^2) T^4}{\rho (\frac{4}{3} \pi r^3) s} = \frac{3 \sigma e T^4}{\rho r s}$.This shows that $\frac{dT}{dt} \propto \frac{1}{r}$.Therefore,$\frac{(\frac{dT}{dt})_P}{(\frac{dT}{dt})_Q} = \frac{r_Q}{r_P} = \frac{3r}{r} = 3$.Given $(\frac{dT}{dt})_P = x (\frac{dT}{dt})_Q$,we find $x = 3$.

Spheres $P$ and $Q$ are uniformly constructed from the same material,which is a good conductor of heat,and the radius of $Q$ is thrice the radius of $P$. The rate of fall of temperature of $P$ is $x$ times that of $Q$ when both are at the same surface temperature. The value of $x$ is:

Similar Questions

$A$ black body is at a temperature $300 K$. It emits energy at a rate,which is proportional to

Find the emissive power from the $0.3 \, cm^2$ surface of a tungsten filament bulb at $3000 \, K$ temperature. Take emissivity $e = 0.4$ for the tungsten bulb. (in $, W$)

When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 \mu m$ to $0.13 \mu m$. The ratio of the emissive powers of the body at the respective temperatures is

$A$ black body is heated from $27^oC$ to $127^oC$. The ratio of their energies of radiations emitted will be

$A$ black body radiates heat energy at the rate of $2 \times 10^5 \, J/s \cdot m^2$ at a temperature of $127 \, ^\circ C$. The temperature of the black body at which the rate becomes $32 \times 10^5 \, J/s \cdot m^2$ is ....... $^\circ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module