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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 places past the decimal in units of kJ (don't include units in your answer).

Respuesta :

Answer:The change in internal energy , [tex]\Delta E[/tex] is 8.09 kJ

Explanation:

According to first law of thermodynamics:

[tex]\Delta E=q+w[/tex]

[tex]\Delta E[/tex]=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) 

[tex]\Delta E=+12.43kJ+(-4.34)=8.09kJ[/tex]

Thus change in internal energy is 8.09 kJ

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