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 $

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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 $



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