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 =10gmMass of liquid in x = 50gmTemperature = 25°CSpecific heat of liquid =12cal/gm°C Water equivalent = 10gmMass of liquid in y= 50gmTemperature = 25°CSpecific heat capacity of liquid =12cal/gm°CMass of A metal substance dropped in calorimeter X=10gmSpecific heat capacity of metal A=0.4 cal/gm°CMass of B metal substance dropped in calorimeter Y=25gmSpecific heat capacity of metal B=? =SBGiven: initial temperature of B is twice the temperature of Aso let the temperature of A be T°C and the temperature of B be 2T°Cand Given: Final temperature of X with A=30°CFinal temperature of Y with B=40°CAs X and A are in thermal equilibrium Y and B are in thermal equilibriumSo using the principle of calorimeter Heat gained by calorimeter = Heat lost by metalSo for now X Heat gained by calorimeterX+ liquid in X = Heat lost by metal A (1)Now firstly we get heat gained by calorimeter XHeat gained by calorimeter is equivalent to heat given to water of 10gm to increase the temperature from 25°C to 30°C =CQ1=mcΔTQ1=10×1×5Q1=50caloriesHeat required to increase the temperature of liquidQ2=mcΔTQ2=50×12×5 Q2=125caloriesTotal heat gained by calorimeter + liquid =Q1+Q2=175caloriesAccording to equation (1)Heat lost by metal A =175caloriesmcΔT=175calories10×0.4×(T−30)=1754T−120=1754T=295T=(2954)0Cso initial temperature of A metal =(2954)0Cand initial temperature of B metal =(2952)0CNow for Y calorimeterHeat 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=150caloriesHeat gained by liquid in Y =Q4=mcΔT=50×12×15=Q4=375caloriestotal heat gained by calorimeter Y + liquid =Q3+Q4=525caloriesUsing the equation (2)Heat lost by metal B = 525 caloriesmcΔT = 525 calories25× SB×(2952−40)=52525×SB×2152=525SB=525×225×215 SB=(42215)cal/gm°C

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