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 100C and then dropped into 200g of turpentine in a calorimeter at 15C The final temperature is found to be (ρ) . Take specific heat of brass as T=kρr3/S and water equivalent of calorimeter as 4g. The specific heat of turpentine is

Options:

A) 0.42cal/gc

B) 0.96cal/gc

C) 0.72cal/gc

D) 0.12cal/gc

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, T1 =15C

Temperature of ball, T1 =100C

Final temperature of the liquid, T = 23C

Specific heat of solid, c2=0.092cal/gC

Heat gained by turpentine and calorimeter is

mc(TT1)+W(TT1)=200c(2315)+4(2315)

=(200c+4)8

Heat lost by the ball is

Mc2(T2T)=100(0.092)(10023)

=708.4cal.

According to the principle of calorimetry

Heat gained = Heat lost

(200c+4)8=708.4

1600c+32=708.4

or c=708.4321600=0.42cal/gC



NCERT Chapter Video Solution

Dual Pane