Transmission Of Heat Question 116
Question: The total energy radiated from a black body source is collected for one minute and is used to heat a quantity of water. The temperature of water is found to increase form $ 20^{o}C $ to $ {{20.5}^{o}}C $ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $ 20^{o}C $ , the temperature of water will be
[UPSEAT 2004]
Options:
A) $ 21^{o}C $
B) $ 22^{o}C $
C) $ 24^{o}C $
D) $ 28^{o}C $
Show Answer
Answer:
Correct Answer: D
Solution:
The total energy radiated from a black body per minute. $ Q\propto T^{4} $
Therefore $ \frac{Q _{2}}{Q _{1}}={{( \frac{2T}{T} )}^{4}}=16 $
Therefore $ Q _{2}=16Q _{1} $ If m be mass of water taken and S be its specific heat capacity, then $ Q _{1}=ms(20.5-20) $ and $ Q _{2}=ms(\theta -20) $
$ \theta {}^\circ C= $ Final temperature of water
Therefore $ \frac{Q _{2}}{Q _{1}}=\frac{\theta -20}{0.5} $
Therefore $ \frac{16}{1}=\frac{\theta -20}{0.5} $
Therefore $ \theta =28{}^\circ C $
BETA
