Question

When 430 J of work was done on a system, it lost 120 J of energy as heat.
Calculate the value of internal energy change (ΔU) for this process.

Answer 1

Answer:

The value of internal energy change is 310 J

Explanation:

According to first law of thermodynamics:

\Delta E=q+wΔE=q+w

\Delta EΔE =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−PΔV {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)=310JΔE=+430+(−120)=310J

Thus the value of internal energy change is 310 J

Learn more about first law of thermodynamics

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Answer 2

Answer:

$= \frac{4}{32000}\times 1000 \times100 = 12.5cm$