Find the external work done by the system in kcal, when 20 kcal of heat is supplied to the system the increase in the internal energy is 8400 J(J=4200J/kcal)
Given,
Q(heat)=20 kcal=20×4200=84000J
ΔU(change in internal energy)=8400J
W=work done
From first law of thermodynamics
Q=ΔU+W
W=Q−ΔU
W=84000−8400=75600 J
W=756004200=18kcal
Work done is positive,
Hence, work done by the system is 18kcal