Properties Of Solids And Liquids Question 46
Question: A water fall is 84 metres high. If half of the potential energy of the falling water gets converted to heat, the rise in temperature of water will be [JIPMER 2002]
Options:
A) 0.098°C
B) 0.98°C
C) 9.8°C
D) 0.0098°C
Show Answer
Answer:
Correct Answer: A
Solution:
As $ W=JQ $ therefore $ \frac{1}{2}(mgh)=J\times mc\Delta \theta $
therefore $ \Delta \theta =\frac{gh}{2JS} $
$ \Delta \theta =\frac{9.8\times 84}{2\times 4.2\times 1000}=0.098{}^\circ C $
$ (\because {S _{\text{water}}}=1000\frac{cal}{kg\times {}^\circ C}) $
Short trick: Remember the value of $ \frac{g}{Jc _{W}}=0.0023 $ ,
here $ \Delta \theta =\frac{1}{2}\times (0.0023)h=\frac{1}{2}\times 0.0023\times 84=0.098{}^\circ C $
