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In a certain process, the energy of the system decreases by 250 kJkJ. The process involves 480 kJkJ of work done on the system. Find the amount of heat QQQ transferred in this process

Answer :

pstnonsonjoku

Answer:

-730KJ

Explanation:

According to the first law of thermodynamics;

Let the total energy of the system be ∆E

Let heat be q and let work the w

Since the energy decreases ∆E is negative

Since work is done on the system w= positive

So;

- 250 = 480 + q

q = -250 - 480

q=-730KJ

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