Question

Thermodynamics :

Answer 1

When the system goes from the state A to the state B, through the trajectory A-C-B, absorbs a heat of 20,000 calories and makes a work of 7,500 calories. a) How much heat absorbs the system through the trajectory A-D-B, if the work done is 2,500 calories
b) When the system returns to the state A from the state B, through the curve, the work done is 5,000 calories. How much heat is absorbed or released?
c) If Ua = 0 cal, and Ud = 10,000 cal, calculate the heat exchanged in the processes A-D and D-B. (U : internal energy)
Answers : a) Q =15,000 cal ; b) Q= -17,500 cal ; c) Qad= 12,500 cal ; d) Qdb = 2,500 cal

Answer 2

Look at the cycle in the attachment

Answer 3

everytime a work is done the energy is negative, and when absorb then it's positive. so what you have is work + absorb

Answer 4

I think in engineering, we assume that the work done is positive

Answer 5

no. for every time a work is done, you are releasing energy, so that is a negative value. the engineering concept is not different from physics or chemistry.

Answer 6

the only time it becomes tricky is when you are dealing with living systems, since living things cannot use heat as energy.

Answer 7

Then, I have to use U = Q - W ???

Answer 8

ya

Answer 9

Ah, I MADE IT !!

Answer 10

first law of thermodynamics

Answer 11

you know, even if you use w + q, you will still get the same answer. that is to make sure that w is negative :D

Answer 12

look at the problem when:
\[\Delta U=Q-W\]
if W is negative then your equation changes to
\[\Delta U=Q-(-W)=Q+W\]
it becomes mathematically incoherent, yes?

Answer 13

THe internal energy variation is equal in the process A-C-B and the process A-D-B
\[\Delta Ucb = Qcb - 7,500 \cal\]
\[\Delta Uac = Qac - 0 (isocoric) \]
\[Qcb = \Delta Ucb + 7,500 \cal\]
\[Qac + Qcb = 20,000 \cal\]
\[\Delta Uac + \Delta Ucb + 7,500 \cal = 20,000 \cal \]
\[\Delta Uac + \Delta Ucb = 12,500 calories\]

\[\Delta Uad + \Delta Udb = 12,500 [calories]\]
\[Qad - 2,500 + Qdb = 12,500 \]
\[Qad + Qdb = 15,000 [calories]\]

Answer 14

I didn't notice the graph LOL!!!

Answer 15

Hahahahah. I didn't remember that when a process is reversible, it only matters the initial and final states, so the internal energy variation in both ways is the same