Transmission Of Heat Question 232
Question: Two identical metal balls at temperature $ 200^{o}C $ and $ 400^{o}C $ kept in air at $ 27^{o}C $ . The ratio of net heat loss by these bodies is
[CPMT 2002]
Options:
A) 1/4
B) 1/2
C) 1/16
D) If temperature of surrounding is considered then net loss of energy of a body by radiation $ Q=A\varepsilon \sigma (T^{4}-T _{0}^{4})t $
Therefore $ Q\propto (T^{4}-T _{0}^{4}) $
Therefore $ \frac{Q _{1}}{Q _{2}}=\frac{T _{1}^{4}-T _{0}^{4}}{T _{2}^{4}-T _{0}^{4}} $ $ =\frac{{{(273+200)}^{4}}-{{(273+27)}^{4}}}{{{(273+400)}^{4}}-{{(273+27)}^{4}}} $ $ =\frac{{{(473)}^{4}}-{{(300)}^{4}}}{{{(673)}^{4}}-{{(300)}^{4}}} $
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Answer:
Correct Answer: C
Solution:
$ \frac{473^{4}-300^{4}}{673^{4}-300^{4}} $
BETA
