Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
753 views
in Physics by (90.2k points)
closed by
A black body with surface area 0.001 `m^(2)` is heated upto a temperature 400 K and is suspended in a room temperature 300K. The intitial rate of loss of heat from the body to room is
A. 10 W
B. 1W
C. 0.1W
D. 0.5 W

1 Answer

0 votes
by (90.9k points)
selected by
 
Best answer
Correct Answer - B
b) The rate of heat emission, u = `sigmaAT^(4)`.
At room temperatures to the rate of absorption is
`u_(0)=sigmaAT_(0)^(4)`
The net rate of heat loss.
`u-u_(0)=sigmaA(T^(4)-T_(0)^(4))`
Given that A=0.001`m^(2)`
T=400K and `T_(0)`=300K
Thus, `u-u_(0)` = `5.67 xx 10^(-8) xx 0.001 xx [(400)^(4)-(300)^(4)]`=0.99 W

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...