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

• Tesla In Person
In summary, 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. The efficiency of the machine is 20%, meaning that 20% of the heat received from the hot source is converted into work. The work done is equal to 1/4 of the heat transferred from the cold source, with a negative sign.
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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 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 / QH . QH here is Q2 and Qc 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

Tesla In Person said:
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.

Tesla In Person said:
2) To solve this I used the equation e = W / QH . QH here is Q2 and Qc 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?

Tesla In Person said:
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.

Tesla In Person and 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

Last edited:
Tesla In Person
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 $$Q_1=-4W$$

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## 1. What is a thermodynamic cycle?

A thermodynamic cycle is a sequence of thermodynamic processes that occur in a closed system, resulting in the system returning to its initial state. This cycle allows for the transfer of energy in the form of heat and work, and is commonly used in engines and power plants.

## 2. How is work done related to heat absorbed in a thermodynamic cycle?

The work done in a thermodynamic cycle is directly related to the heat absorbed by the system. This is known as the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

## 3. What is the efficiency of a thermodynamic cycle?

The efficiency of a thermodynamic cycle is a measure of how much of the energy input is converted into useful work. It is calculated by dividing the work output by the heat input, and is always less than 100% due to the second law of thermodynamics.

## 4. How does the Carnot cycle relate to thermodynamic cycles?

The Carnot cycle is a theoretical thermodynamic cycle that is considered to be the most efficient possible cycle for converting heat into work. It serves as a benchmark for comparing the efficiency of other thermodynamic cycles and is based on the principles of reversible processes and the Carnot theorem.

## 5. What are some real-world applications of thermodynamic cycles?

Thermodynamic cycles are used in many real-world applications, such as in power plants, refrigeration systems, and engines. They are also used in the study of atmospheric phenomena, such as weather patterns and climate change. Additionally, thermodynamic cycles are used in the design and optimization of various industrial processes, such as chemical reactions and manufacturing processes.

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