The total energy radiated from a black body s | vedclass

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

Question

(D) According to the Stefan-Boltzmann law,the total energy $Q$ radiated from a black body per unit time is proportional to the fourth power of its abs

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) According to the Stefan-Boltzmann law,the total energy $Q$ radiated from a black body per unit time is proportional to the fourth power of its absolute temperature $T$,i.e.,$Q \propto T^4$.Let $Q_1$ be the energy radiated at temperature $T$ and $Q_2$ be the energy radiated at temperature $2T$.Then,$\frac{Q_2}{Q_1} = \left( \frac{2T}{T} ight)^4 = 2^4 = 16$.So,$Q_2 = 16 Q_1$.The heat absorbed by water is given by $Q = ms\Delta	heta$,where $m$ is the mass of water and $s$ is the specific heat capacity.For the first case: $Q_1 = ms(20.5 - 20) = ms(0.5)$.For the second case: $Q_2 = ms(	heta - 20)$,where $	heta$ is the final temperature.Dividing the two equations:$\frac{Q_2}{Q_1} = \frac{ms(	heta - 20)}{ms(0.5)} = \frac{	heta - 20}{0.5}$.Since $\frac{Q_2}{Q_1} = 16$,we have:$16 = \frac{	heta - 20}{0.5}$.$8 = 	heta - 20$.$	heta = 28^{\circ}C$.

Similar Questions

The temperature of a black body increases from $327^{\circ}C$ to $927^{\circ}C$. If the initial energy possessed is $2 \ kJ$,what is its final energy in $kJ$?

If the temperature of the sun is doubled,the rate of energy received by the earth will be increased by a factor of:

Two spheres of radii $4 \, m$ and $1 \, m$ have temperatures $2000 \, K$ and $4000 \, K$ respectively. Find the ratio of their emitted energy.

The temperatures of two bodies $A$ and $B$ are $727^{\circ}C$ and $327^{\circ}C$ respectively. What is the ratio of their rates of energy emission $H_A : H_B$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module