Q.6 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)
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Answer:
∆u=6.4×10^4J
Step-by-step explanation:
given:Q=20kcl, W=20000J, J=4200J/kcl
formula:∆u=|Q|-|W|
to calculate: ∆u=?
calculations: convert the given heat from kcal to joule.
Q(in joule) = J×Q(in kcal)
=4200×20
=84000J
therefore, ∆u=|Q|-|w|
=84000-20000
=64000
=6.4×10^4J
*the increase in the internal energy of the system is 6.4×10^4J
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