Q. 15.0( 1 Vote )

A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J kg–1 K–1 and 4200 J kg–1 K–1 respectively.

Neglect any thermal expansion.

(a) How much heat is transferred to the liquid-vessel system?

(b) How much work has been done on this system?

(c) How much is the increase in internal energy of the system?

Answer :

Given


Initial temperature of water T1=15 =288K


Final temperature of water T2=17 =290K


Specific heat capacity of copper cc=420 J kg–1 K–1


Specific heat capacity of water cw=4200 J kg–1 K–1


Mass of copper vessel mc=100g = 100×10-3kg


Mass of water mw=200g = 200×10-3kg


a) It is given that copper vessel is thermally insulated. Therefore, no heat from the surroundings can be transferred to the liquid-vessel system.


b) Work done on this system will be


ΔW=mwcwΔT + mcccΔT


ΔT=T2-T1=290-288=2K


So,


ΔW= 200×10-3×4200×2 + 100×10-3×420×2


ΔW=1680 + 84=1764J


work done on the system is 1764J.


c) From first law of thermodynamics, we know that,


ΔQ=ΔU+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


From part (a) we have concluded that ΔQ=0


So, first law becomes


ΔU=-ΔW


Work done on the system ΔW=1764 J


So, work done by the system ΔW=-1764J


ΔU = -(-1764) = 1764J


Increase in internal energy of the system = 1764J.


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