Q. 125.0( 1 Vote )

A substance is taken through the process abc as shown in the figure. If the internal energy of the substance increases by 5000 J and heat of 2625 cal is given to the system, calculate the value of J.



Answer :

‘J' is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to the joule.


Given


Heat given to system =2625cal =2625×J J


Change in internal energy= 5000J


From graph


Va=0.02m3


Vb=Vc=0.05m3


Pa=Pb=200kPa=200×103Pa


Pc=300kPa=300×103Pa


We know that work done by the gas is given as


ΔW=PΔV


Work done in process abc=ΔW=Wab+Wbc


ΔW=Pa(Vb-Va)+0 ( Wbc=0 because Vb=Vc)


ΔW=200×103×(0.05-0.02)=6000J


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


ΔQ=5000+6000=11000J


But ΔQ=2625cal =2625×J J


Therefore,


2625×J=11000



value of ‘J’ is 4.19joule/cal.


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