Question

When 20 kcal heat is supplied to a system, the
external work done is 20,000 J. Find the
increase in integral energy of the system (in
joule) (J=4200 J/kcal)​

Answer 1

Answer:

Correct option is

B

18 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=

4200

75600

=18kcal

Work done is positive,

Hence, work done by the system is 18kcal