Transmission Of Heat Question 242
Question: At 127o C radiates energy is 2.7 ´ 10-3 J/s. At what temperature radiated energy is 4.32 ´ 10 6 J/s
[BCECE 2004]
Options:
A) 400 K
B) 4000 K
C) 80000 K
D) 40000 K
Show Answer
Answer:
Correct Answer: C
Solution:
Energy radiated from a body $ Q=A\varepsilon \sigma T^{4}t $
Therefore $ \frac{Q _{2}}{Q _{1}}={{( \frac{T _{2}}{T _{1}} )}^{4}} $
Therefore $ \frac{T _{2}}{T _{1}}={{( \frac{Q _{2}}{Q _{1}} )}^{1/4}}={{( \frac{4.32\times 10^{6}}{2.7\times {{10}^{-3}}} )}^{1/4}} $
$ ={{( \frac{16\times 27}{27}\times 10^{8} )}^{1/4}} $
$ =2\times 10^{2} $
Therefore $ T _{2}=200\times T _{1}=80000,K $
BETA
