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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, 25C and 12calg1C1 respectively. A 10 g metal piece (say A) of specific heat capacity 0.4calg1C1 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 30C and 40C 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=? =SB
Given: 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×(T30)=175
4T120=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×(295240)=525
25×SB×2152=525
SB=525×225×215
SB=(42215)cal/gm°C

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