Question

12)The internal enery of the system is 560 and the work done by surroundings is 320....find

Q...?​

Answer 1

Answer:

HY MATE HERE IS YOUR ANSWER

From first law of thermodynamic, Q=ΔU+W

From first law of thermodynamic, Q=ΔU+WAs the work done on the system so it will be negative i.e, W=−320J and since internal energy is increased so it would be positive i.e, ΔU=560J

From first law of thermodynamic, Q=ΔU+WAs the work done on the system so it will be negative i.e, W=−320J and since internal energy is increased so it would be positive i.e, ΔU=560JThus, dQ=560−320=240J 

From first law of thermodynamic, Q=ΔU+WAs the work done on the system so it will be negative i.e, W=−320J and since internal energy is increased so it would be positive i.e, ΔU=560JThus, dQ=560−320=240J As Q is positive so the system will absorb heat.  

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Answer 2

Answer:

In the earth internal enery of the system is 560 and the work done by surroundings is 320 then its answer is 560 +320= 880