A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy

Question

A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy (A) increases by 600 J (B) decreases by 800 J (C) increases by 800 J (D) decreases by 50 J

Tardigrade Admin · Accepted Answer

dQ=+ 200 cal =200× 4.2=840 J dW=+ 40 J From the first law of thermodynamics dQ=dU+dW dU=dQ-dW =840-40=800 J So, the internal energy of the .system increased by 800 J.

Q. A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy

increases by 600 J

decreases by 800 J

increases by 800 J

decreases by 50 J

$d Q = + 200 c a l$ $= 200 \times 4.2 = 840 J$ $d W = + 40 J$ From the first law of thermodynamics $d Q = d U + d W$ $d U = d Q - d W$ $= 840 - 40 = 800 J$ So, the internal energy of the .system increased by 800 J.

Q. A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy

increases by 600 J

decreases by 800 J

increases by 800 J

decreases by 50 J

dQ=+200cal =200×4.2=840J dW=+40J From the first law of thermodynamics dQ=dU+dW dU=dQ−dW =840−40=800J So, the internal energy of the .system increased by 800 J.

$d Q = + 200 c a l$ $= 200 \times 4.2 = 840 J$ $d W = + 40 J$ From the first law of thermodynamics $d Q = d U + d W$ $d U = d Q - d W$ $= 840 - 40 = 800 J$ So, the internal energy of the .system increased by 800 J.