Transmission Of Heat Question 118
Question: A solid copper cube of edges $ 1\ cm $ is suspended in an evacuated enclosure. Its temperature is found to fall from $ 100^{o}C $ to $ 99^{o}C $ in $ 100\ s $ . Another solid copper cube of edges $ 2\ cm $ , with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from $ 100^{o}C $ to $ 99^{o}C $ will be approximately
[MP PMT 1997]
Options:
A) $ 25\ s $
B) $ 50\ s $
C) $ 200\ s $
D) $ 400\ s $
Show Answer
Answer:
Correct Answer: C
Solution:
Rate of cooling $ \frac{\Delta \theta }{t}=\frac{A\varepsilon \sigma (T^{4}-T _{0}^{4})}{mc} $
Therefore $ t\propto \frac{m}{A} $
$ [\because \ \ \Delta \theta ,\ t,\ \sigma ,\ \ (T^{4}-T _{0}^{4})\ \text{are}\ \text{constant}] $
Therefore $ t\propto \frac{m}{A}\propto \frac{\text{Volume}}{\text{Area}}\propto \frac{a^{3}}{a^{2}} $
Therefore $ t\propto a $
Therefore $ \frac{t _{1}}{t _{2}}=\frac{a _{1}}{a _{2}} $
Therefore $ \frac{100}{t _{2}}=\frac{1}{2} $
Therefore $ t _{2}=200sec. $
BETA
