Question

When 430 joule of work was done on a system at lost 120 joule of energy acid calculate the value of internal energy change?

Answer 1

Answer:

310 J

Explanation:

∆U=q+(-w)

for this case q is negative

Answer 2

The value of internal energy change is 310 J

Explanation:

According to first law of thermodynamics:

$\Delta E=q+w$

$\Delta E$ =Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system= $-P\Delta V$ {Work is done on the system is positive as the final volume is lesser than initial volume}

w = +430 J

q = -120 J   {Heat released by the system is negative}

$\Delta E=+430+(-120)=310J$

Thus the value of internal energy change is 310 J

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