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AleksAgata [21]
4 years ago
6

Calculate ΔE for a process in which a system performs 4.34 kJ of work and absorbs 12.43 kJ of heat. Express your answer to two p

laces past the decimal in units of kJ (don't include units in your answer).
Physics
1 answer:
Anettt [7]4 years ago
5 0

Answer:The change in internal energy , \Delta E is 8.09 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released  = +12.43 kJ  (heat absorbed is positivew = work done or by the system  = - 4.34 kJ (Work done by the system is negative) 

\Delta E=+12.43kJ+(-4.34)=8.09kJ

Thus change in internal energy is 8.09 kJ

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Explanation:

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See the picture below

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