110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the amount of external work done is
Correct answer: B. 70 J
- A. 150 J
- B. 70 J
- C. 110 J
- D. 40 J
Explanation
Correct. Applying the first law of thermodynamics: ΔU = Q - W W = Q - ΔU Q = 110 J ΔU = 40 J W = 110 J - 40 J = 70 J
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About First Law of Thermodynamics
The first law relates heat supplied, work done and the change in internal energy through energy conservation. Problems use sign conventions and apply the law to isothermal, adiabatic, isobaric and isochoric processes. Internal energy is a state function, while heat and work depend on the path followed.
Practise Thermodynamics
714 free Thermodynamics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More First Law of Thermodynamics questions
Find the change in internal energy of the system when a system absorbs 2 kilocalorie of heat and at the same time does 500 joule of work
If heat equal to 0.1 J is provided to the gas contained in a cylinder and it expands through 0.1m3 at 1N/m2 then its internal energy:
Carnot cycle consists of
When the amount of work done is 333 cal and change in internal energy is 167 cal, then the heat supplied is
If the heat absorbed is 10 J and work done is 5 J, then the change in internal energy is:
In which process is the net work done zero?