Question

Two identical calorimeters 'X' and 'Y', of water equivalent 10 g each, contain equal quantities of the same liquid. The mass, initial temperature and specific heat capacity of the liquid present in both calorimeters are 50 g, 25∘C and 12calg−1∘C−1 respectively. A 10 g metal piece (say A) of specific heat capacity 0.4calg−1∘C−1 is dropped in calorimeter "X". and 25 g of metal piece (say B) of specific heat capacity "SB" is dropped into calorimeter "Y". Due to this, the equilibrium temperature in 'X' and 'Y' rises to 30∘C and 40∘C respectively. If the initial temperature of 'B' is twice that of 'A', then find the value of "SB".

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Solution

X | Y |

Water equivalent =10gm Mass of liquid in x = 50gm Temperature = 25°C Specific heat of liquid =12cal/gm°C | Water equivalent = 10gm Mass of liquid in y= 50gm Temperature = 25°C Specific heat capacity of liquid =12cal/gm°C |

Mass of A metal substance dropped in calorimeter X=10gm

Specific heat capacity of metal A=0.4 cal/gm°C

Mass of B metal substance dropped in calorimeter Y=25gm

Specific heat capacity of metal B=? =SBGiven: initial temperature of B is twice the temperature of A

so let the temperature of A be T°C

and the temperature of B be 2T°C

and Given: Final temperature of X with A=30°C

Final temperature of Y with B=40°C

As X and A are in thermal equilibrium

Y and B are in thermal equilibrium

So using the principle of calorimeter

Heat gained by calorimeter = Heat lost by metal

So for now X

Heat gained by calorimeterX+ liquid in X = Heat lost by metal A (1)

Now firstly we get heat gained by calorimeter X

Heat gained by calorimeter is equivalent to heat given to water of 10gm to increase the temperature from 25°C to 30°C =C

Q1=mcΔT

Q1=10×1×5

Q1=50calories

Heat required to increase the temperature of liquid

Q2=mcΔT

Q2=50×12×5

Q2=125calories

Total heat gained by calorimeter + liquid =Q1+Q2=175calories

According to equation (1)

Heat lost by metal A =175calories

mcΔT=175calories

10×0.4×(T−30)=175

4T−120=175

4T=295

T=(2954)0C

so initial temperature of A metal =(2954)0C

and initial temperature of B metal =(2952)0C

Now for Y calorimeter

Heat gained by calorimeter Y + liquid = Heat lost by metal B (2)

Now heat gained by calorimeter = heat given to water of 10gm to increase the temperature from25°C to 45°C =Q3

=Q3=mcΔT

=Q3=10×1×15

=Q3=150calories

Heat gained by liquid in Y

=Q4=mcΔT

=50×12×15

=Q4=375calories

total heat gained by calorimeter Y + liquid =Q3+Q4=525calories

Using the equation (2)

Heat lost by metal B = 525 calories

mcΔT = 525 calories

25× SB×(2952−40)=525

25×SB×2152=525

SB=525×225×215

SB=(42215)cal/gm°C

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