- #1

Tesla In Person

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- Homework Statement
- Work done as a function of Heat absorbed

- Relevant Equations
- e = W / Qh

**During a thermodynamic cycle, an ideal thermal machine absorbs heat Q**I have 2 question regarding this problem: 1) Why is Q

_{2}> 0 from a hot source and uses it to perform Work W > 0, giving a cold source a heat Q_{1}< 0 with an efficiency of 20% . How much is the work done as a function of Q_{1}?_{1}the heat given to cold source negative? Is it because it leaves the system ?

2) To solve this I used the equation e = W / Q

_{H}. Q

_{H}here is Q

_{2}and Q

_{c}is Q

_{1}. Also

Q

_{2}= Q

_{1}+ W . So substituting this into the first equation it becomes e = W / (Q

_{1}+ W ) setting this equation equal to 20 which is the efficiency and simplifying gives the answer W = 1/4 Q

_{1}but it's wrong the right answer has a negative sign. It's - Q

_{1}/ 4 . Can anyone explain why ? Thanks