The temperature of a blackbody radiation encl | vedclass

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

Question

(D) The energy density $u$ of blackbody radiation at temperature $T$ is given by $u = aT^4$, where $a$ is the radiation constant. The total energy $E$

Vedclass · Accepted Answer

Information not sufficient

Vedclass · Answer

(D) The energy density $u$ of blackbody radiation at temperature $T$ is given by $u = aT^4$, where $a$ is the radiation constant. The total energy $E$ in a container of volume $V$ is $E = uV = aVT^4$. The heat required to increase the temperature from $T_1$ to $T_2$ is $\Delta E = aV(T_2^4 - T_1^4)$. Since the volume $V$ of the container is not provided in the problem statement, it is impossible to calculate the exact numerical value of the heat required. Therefore, the information provided is insufficient.

The temperature of a blackbody radiation enclosed in a container of volume $V$ is increased from $100^{\circ} C$ to $1000^{\circ} C$. The heat required in the process is

Similar Questions

Define emissivity and absorptivity and give their values for a perfect black body.

If a vessel with black spots is heated to a very high temperature and then taken into a dark room,then:

Three objects colored black,gray,and white can withstand hostile conditions up to $2800\,^oC$. These objects are thrown into a furnace where each of them attains a temperature of $2000\,^oC$. Which object will glow brightest?

For a perfectly black body,the coefficient of emission (emissivity) is:

In which process is the rate of transfer of heat maximum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module