Work greater than heat given to colder reservoir -- impossible?

[Callmelucky]

Homework Statement

Is it possible to make machine that would take more heat from hot reservoir to do work than what would hot reservoir give to cold reservoir(heat)?

Relevant Equations

1st law of thermodynamics

photo below... Is it possible to make a machine that would take more heat from the hot reservoir to do work than what would hot reservoir give to the cold reservoir(heat)? Apparently, it's impossible because it violates 1st law of thermodynamics.

the thickness of the arrows symbolizes the amount of heat or work. The red rectangle is hot res. the blue rectangle is cold res. The grey circle is a working tool(mean).

 

[Drakkith]

I'm unsure of what you're asking. Are you asking about the total amount of heat (or energy) moved, the rate of heat transfer, or something else?

 

[Callmelucky]

I'm unsure of what you're asking. Are you asking about the total amount of heat (or energy) moved, the rate of heat transfer, or something else?

Apparently, heat that goes from a hot reservoir to be used to do work can not be greater than the heat that goes to a cold reservoir, is that true?

 

[haruspex]

Apparently, heat that goes from a hot reservoir to be used to do work can not be greater than the heat that goes to a cold reservoir, is that true?

Seems you have it backwards. As the diagram illustrates, it is certainly possible that $Q_1>Q_2$. It is the reverse that is not possible.

 

[Callmelucky]

Seems you have it backwards. As the diagram illustrates, it is certainly possible that $Q_1>Q_2$. It is the reverse that is not possible.

I understand that, I don't understand why(according to answers at the end of the book) work can not be greater that Q2

 

[Drakkith]

I understand that, I don't understand why(according to answers at the end of the book) work can not be greater that Q2

As far as I know it can. GE has a gas turbine that runs at upwards of 64% efficiency, meaning that W > Q2.

SCHENECTADY, NEW YORK - December 4, 2017 - GE Power (NYSE:GE) announced today that its largest and most efficient gas turbine, the HA, is now available at more than 64 percent efficiency in combined cycle power plants, higher than any other competing technology today. This milestone is achieved largely due to GE’s advances in additive manufacturing and combustion breakthroughs through constant innovation, coming just 18 months after GE set a world record with the HA for powering the world’s most efficient combined-cycle power plant at 62.22 percent.

Link here.

 

[Drakkith]

Also, the theoretical maximum efficiency of a heat engine is given by: $n_{max}=1-\frac{T_c}{T_h}$
So there's nothing theoretically stopping a heat engine from being more than 50% efficient as long as the temperature difference is high enough.

 

[256bits]

Also, the theoretical maximum efficiency of a heat engine is given by: $n_{max}=1-\frac{T_c}{T_h}$
So there's nothing theoretically stopping a heat engine from being more than 50% efficient as long as the temperature difference is high enough.

OP said according to the 1st law, of which he has not explained how it bears upon the scenario he has described.
@Callmelucky Could you explain your reasoning with regards to the 1st law?
drakkith has proposed the 2nd law.

 

[haruspex]

I understand that, I don't understand why(according to answers at the end of the book) work can not be greater that Q2

I may have misunderstood. Is it saying W cannot exceed Q₂? Seems to me it can, provided $T_1>2T_2$.

 

[Callmelucky]

the question goes: is this heat engine possible? if not, what law of thermodynamics is he violating? And in solutions, the answer is: not possible, violates 1st law of thermodynamics. Which is weird because in that same textbook, in the chapter on 1st law of thermodynamics it's stated that some ship engines have efficiency of 52%. I guess it's a mistake then in solutions.

 

[erobz]

The work cannot equal $Q_1$ by the Kelvin-Plank statement of the Second Law, but that's not what is portrayed.

I think your book is incorrect. What's shown seems quite plausible for any real process.

 

[Callmelucky]

The work cannot equal $Q_1$ by the Kelvin-Plank statement of the Second Law, but that's not what is portrayed.

I think your book is incorrect. What's shown seems quite plausible for any real process.

Yes, something happend in solutions, I think that the order of answers is wrong and some are missing.
My bad for not checking them all first, I'm sorry for that.

 

[erobz]

Yes, something happend in solutions, I think that the order of answers is wrong and some are missing.
My bad for not checking them all first, I'm sorry for that.

I'm no expert. Don't be sorry, these things can be subtle ( that's why perpetual motion devices are "a thing"). You shouldn't have to check the book. However, It's better that you first questioned your understanding and then confirmed your suspicions IMHO.

 

[kuruman]

We are told that "the thickness of the arrows symbolizes the amount of heat or work." I used a plastic ruler to measure the width of each arrow on my laptop's screen. I got
Q₁ = 1.9 ± 0.05 cm (total heat entering the radno sredstvo from the red, hi T, reservoir)
Q₂ = 0.6 ± 0.05 cm (total heat dumped by the radno sredstvo into the blue, lo T, reservoir)
W = 1.2 ± 0.05 cm (work done by the radno sredstvo)

The first law says Q₁ - Q₂ = W over one cycle.
Here Q₁ - Q₂ = 1.3 ± 0.1 cm.
My conclusion is that, to within the uncertainty of my measurement, the diagram is consistent with the first law of thermodynamics. The efficiency of the cycle can also be estimated, $$e=\frac{W}{Q_1}=\frac{1.2}{1.9}=0.63\pm 0.03.$$

 

[erobz]

We are told that "the thickness of the arrows symbolizes the amount of heat or work." I used a plastic ruler to measure the width of each arrow on my laptop's screen. I got
Q₁ = 1.9 ± 0.05 cm (total heat entering the radno sredstvo from the red, hi T, reservoir)
Q₂ = 0.6 ± 0.05 cm (total heat dumped by the radno sredstvo into the blue, lo T, reservoir)
W = 1.2 ± 0.05 cm (work done by the radno sredstvo)

The first law says Q₁ - Q₂ = W over one cycle.
Here Q₁ - Q₂ = 1.3 ± 0.1 cm.
My conclusion is that, to within the uncertainty of my measurement, the diagram is consistent with the first law of thermodynamics. The efficiency of the cycle can also be estimated, $$e=\frac{W}{Q_1}=\frac{1.2}{1.9}=0.63\pm 0.03.$$

Well, that should settle it!

 

[kuruman]

Well, that should settle it!

When in doubt, measure it out.

 

[Callmelucky]

We are told that "the thickness of the arrows symbolizes the amount of heat or work." I used a plastic ruler to measure the width of each arrow on my laptop's screen. I got
Q₁ = 1.9 ± 0.05 cm (total heat entering the radno sredstvo from the red, hi T, reservoir)
Q₂ = 0.6 ± 0.05 cm (total heat dumped by the radno sredstvo into the blue, lo T, reservoir)
W = 1.2 ± 0.05 cm (work done by the radno sredstvo)

The first law says Q₁ - Q₂ = W over one cycle.
Here Q₁ - Q₂ = 1.3 ± 0.1 cm.
My conclusion is that, to within the uncertainty of my measurement, the diagram is consistent with the first law of thermodynamics. The efficiency of the cycle can also be estimated, $$e=\frac{W}{Q_1}=\frac{1.2}{1.9}=0.63\pm 0.03.$$

yeah, it's some kind of mistake in textbook. That was a clever thing to do, it didn't come to me because I am pretty sure that it was not supposed to be done that way but yeah, still good idea. Thank you.

 

[kuruman]

yeah, it's some kind of mistake in textbook. That was a clever thing to do, it didn't come to me because I am pretty sure that it was not supposed to be done that way but yeah, still good idea. Thank you.

The only way that physics problems are supposed to be done is correctly.

 

[Lnewqban]

Please, see:
https://en.wikipedia.org/wiki/Carnot_heat_engine