Properties Of Solids And Liquids Question 532
Question: DIRECTION: Read the passage given below and answer the questions that follows:
A brass ball of mass 100g is heated to $ 100{}^\circ C $ and then dropped into 200g of turpentine in a calorimeter at $ 15{}^\circ C $ The final temperature is found to be $ (\rho ) $ . Take specific heat of brass as $ T=k\sqrt{\rho r^{3}/S} $ and water equivalent of calorimeter as 4g. The specific heat of turpentine is
Options:
A) $ 0.42cal/g{}^\circ c $
B) $ 0.96cal/g{}^\circ c $
C) $ 0.72cal/g{}^\circ c $
D) $ 0.12cal/g{}^\circ c $
Show Answer
Answer:
Correct Answer: A
Solution:
Let c be the specific heat of turpentine
Mass of the solid, M = 100g
Mass of turpentine m = 200g
Water equivalent of calorimeter, W = 4g
Initial temperature of calorimeter, $ T{{} _{1}}~=15{}^\circ C $
Temperature of ball, $ T _{1}~=100{}^\circ C $
Final temperature of the liquid, T = $ 23{}^\circ C $
Specific heat of solid, $ c _{2}=0.092cal/{{g}^{{}^\circ }}C $
Heat gained by turpentine and calorimeter is
$ mc( T-T _{1} )+W( T-T _{1} )=200c( 23-15 )+4( 23-15 ) $
$ =( 200c+4 )8 $
Heat lost by the ball is
$ Mc _{2}( T _{2}-T )=100( 0.092 )( 100-23 ) $
$ =708.4cal. $
According to the principle of calorimetry
Heat gained = Heat lost
$ \therefore $ $ ( 200c+4 )8=708.4 $
$ 1600c+32=708.4 $
or $ c=\frac{708.4-32}{1600}=0.42cal/{{g}^{{}^\circ }}C $
