Question

$\small\fbox\pink{ig missdemanding}$ Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the system so that its volume decreases at constant pressure.What is change in its internal energy?

Answer 1

Explanation:

Have Given :-

Heat = 30

external work = 4200 J

ΔQ = 30 × 4200

= 126000J

$ΔW=−4200J$

$∴Δu=ΔQ−ΔW$

$=126000+4200J$

$=1.302×10^{5} J$

I hope it is helpful

Answer 2

Heat = 30

external work = 4200 J

ΔQ = 30 × 4200

= 126000J

ΔW=−4200J

∴Δu=ΔQ−ΔW

=126000+4200J

=1.302×10^{5} J

hopefully its helpful for you :)