when 20 kcal of heat is supplied to the system and the increases in the internal energy is 8400j then work done by is
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Step-by-step explanation:
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= 4200/75600
=18kcal
Work done is positive,
Hence, work done by the system is 18 kacl
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