Transmission Of Heat Question 234
Question: A black body at a temperature of 127°C radiates heat at the rate of 1 cal/cm2 ´ sec. At a temperature of 527°C the rate of heat radiation from the body in (cal/cm2 ´ sec) will be
[MP PET 2002]
Options:
A) 16.0
B) 10.45
C) 4.0
D) 2.0
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{Q _{2}}{Q _{1}}=\frac{T _{2}^{4}}{T _{1}^{4}}={{( \frac{273+527}{273+127} )}^{4}}={{( \frac{800}{400} )}^{4}} $
Therefore $ Q _{2}=16\frac{cal}{cm^{2}\times s} $
BETA
