Thermodynamic Cycle -- Work done as a function of Heat absorbed

[Tesla In Person]

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 Q₂ > 0 from a hot source and uses it to perform Work W > 0, giving a cold source a heat Q₁ < 0 with an efficiency of 20% . How much is the work done as a function of Q₁ ?I have 2 question regarding this problem: 1) Why is Q₁ 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 Q₂ and Q_c is Q₁ . Also
Q₂ = Q₁ + W . So substituting this into the first equation it becomes e = W / (Q₁ + W ) setting this equation equal to 20 which is the efficiency and simplifying gives the answer W = 1/4 Q₁ but it's wrong the right answer has a negative sign. It's - Q₁ / 4 . Can anyone explain why ? Thanks

 

[TSny]

During a thermodynamic cycle, an ideal thermal machine absorbs heat Q₂ > 0 from a hot source and uses it to perform Work W > 0, giving a cold source a heat Q₁ < 0 with an efficiency of 20% . How much is the work done as a function of Q₁ ?1) Why is Q₁ the heat given to cold source negative? Is it because it leaves the system ?

I think the problem statement is incorrect. When a thermal machine takes in positive heat from a hot source and does a positive amount of work, then a positive amount of heat is transferred to the cold source. So, if Q₁ is defined as the heat given to the cold source, then Q₁ is a positive quantity.

If, however, you were to define Q₁ as the heat transferred to the machine from the cold source, then Q₁ would be a negative quantity.

The wording of the problem implies that Q₁ is the heat transferred to the cold reservoir. So, Q₁ should be positive. Thus, it seems to me that the problem statement is contradictory.

2) To solve this I used the equation e = W / Q_H . Q_H here is Q₂ and Q_c is Q₁ . Also
Q₂ = Q₁ + W .

How are you defining Q₁ here? Is it the heat transferred from the machine to the cold source or is it the heat transferred from the cold source to the machine?

So substituting this into the first equation it becomes e = W / (Q₁ + W ) setting this equation equal to 20 which is the efficiency and simplifying gives the answer W = 1/4 Q₁ but it's wrong the right answer has a negative sign. It's - Q₁ / 4 . Can anyone explain why ? Thanks

Whether the answer should be W = 1/4 Q₁ or W = -1/4 Q₁ will depend on how Q₁ is defined.

 

[Steve4Physics]

I agree with @TSny. The wording in the question is poor/misleading.

Edited - correction.

$Q_2$ and $Q_1$ are (probably) meant to be the amounts of heat transferred to and from the machine respectively. The sign-convention being used would then be:

heat entering the machine is positive​

heat leaving the machine is negative​

With this sign-convention, conservation of energy gives:
$Q_2 = W – Q_1$
as used here for example: https://okulaer.files.wordpress.com/2014/08/carnot.jpg

 

[Chestermiller]

In the first law of thermodynamics, Q is the heat transferred from the surroundings (in this case the reservoirs) to the system (in this case, the working fluid of the engine). So, if heat is transferred from the system to the surroundings, it is negative (in first law parlance). So, for this problem, $$W=Q_1+Q_2$$The efficiency is defined as the work done divided by the heat received from the hot reservoir: $$e=\frac{W}{Q_2}=0.2$$So $$Q_2=5W$$and $$Q_1=-4W$$