Carnot Engine question- quick check

In summary, a Carnot heat engine operates between a hot reservoir of boiling water and a cold reservoir of ice and water. In 5 minutes, the engine melts 2.85×10−2 Kg of ice, performing 9519 Joules of work. The maximum efficiency of the engine is 0.27. To solve for the heat rejected by the engine, the equation n = 1 - (Qc/Qh) can be rearranged to find Qh = Qc/(1-n). This can then be used to find the work performed by the engine, which is equal to Qh-Qc.
  • #1
vorcil
398
0
A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water.
In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 2.85×10−2 Kg.

Throughout this problem use Lf=3.34*10^5 j/kg for the heat of fusion for water.

- During this time, how much work is performed by the engine?


my attempt:
i noticed that the two temperature differences are 100 degrees and 0 degrees(mixture of icenwater)
so the maximum efficiency is n = 1- qc/qh = 0.27

how much work is performed by the engine,
Q(work)=Mass melted * latent heat fusion thing
Qc= m*Lf
so (2.85*10^-2 ) * (3.34*10^5) = 9519 Joules

now this is where my question comes in,
do i multiply the efficiency, 0.27 by 9519
to get the work performed by the engine
or is that the work lost by the engine so 9519-.27*9519

i just wanted to check before i submit the awnser(it's mastering physics)
 
Last edited:
Physics news on Phys.org
  • #2
vorcil said:

now this is where my question comes in,
do i multiply the efficiency, 0.27 by 9519
to get the work performed by the engine
or is that the work lost by the engine so 9519-.27*9519

i just wanted to check before i submit the awnser(it's mastering physics)
Go back to the definition of efficiency: W/Qh. In any engine, W = Qh-Qc.

[tex]\eta = 1 - \frac{Q_c}{Q_h}[/tex]

You have Qc. What you want to find first is Qh.

AM
 
  • #3
Andrew Mason said:
Go back to the definition of efficiency: W/Qh. In any engine, W = Qh-Qc.

[tex]\eta = 1 - \frac{Q_c}{Q_h}[/tex]

You have Qc. What you want to find first is Qh.

AM

so Qh = (1 - Qc) / n ?
 
  • #4
vorcil said:
so Qh = (1 - Qc) / n ?

Your algebra is bad.

Qh = (Qh - Qc) / n
 
  • #5
djeitnstine said:
Your algebra is bad.

Qh = (Qh - Qc) / n

I'm sorry, i had to teach myself everything until i got to university,
can you please explain how you solved for Qh, from n = 1-(qc/qh)
so that i might not make that mistake again?

and from this,
Qh = (Qh - Qc) / n
aren't i supposed to know qh to solve for qh? or is that equation the same as n=1-(qh/qc)
 
Last edited:
  • #6
I used the solver in my calculator to find Qh from n = 1 - (qc/qh)
and it was 13039J
w = qh-qc,
13039 - 9519 = 3507 J which was correct
but would still like to know how to rearrange, n = 1 - (qc/qh) to solve for qh =]
thanks everyone
 
  • #7
Your best friend is some pen and paper man...

Flip some numbers around and do some mathematical acrobatics to achieve [tex]Q_h = \frac{Q_c}{1-\eta}[/tex]
 
  • #8
vorcil said:
I used the solver in my calculator to find Qh from n = 1 - (qc/qh)
and it was 13039J
w = qh-qc,
13039 - 9519 = 3507 J which was correct
but would still like to know how to rearrange, n = 1 - (qc/qh) to solve for qh =]
thanks everyone
Vorcil. You need to study algebra before you can hope to do anything here. Get a tutor - fast. It is not difficult but you need this basic skill. When you have an equation, you can rearrange it by doing the same thing to each side:

[tex]\eta = 1 - \frac{Q_c}{Q_h}[/tex]

subtract 1 from both sides and multiply both sides by -1:

[tex]\frac{Q_c}{Q_h} = \eta - 1[/tex]

Multiply both sides by Qh and divide by [itex]\eta - 1[/itex]

[tex]\frac{Q_c}{\eta -1} = Q_h[/tex]

AM
 
  • #9
Andrew Mason said:
Vorcil. You need to study algebra before you can hope to do anything here. Get a tutor - fast. It is not difficult but you need this basic skill. When you have an equation, you can rearrange it by doing the same thing to each side:

[tex]\eta = 1 - \frac{Q_c}{Q_h}[/tex]

subtract 1 from both sides and multiply both sides by -1:

[tex]\frac{Q_c}{Q_h} = \eta - 1[/tex]

Multiply both sides by Qh and divide by [itex]\eta - 1[/itex]

[tex]\frac{Q_c}{\eta -1} = Q_h[/tex]


AM

thanks for showing me, had no idea that you could solve it like that,

as for studying calculus and algebra, i have reasonable skills, I'm doing first year math and physics papers and don't find anything hard except for the occasional weird algebraic things that come up, I've mastered a majority of the basic algebraic skills needed to solve problems in physics but there are some rules that I've never seen before,

for the example above, i knew you could subtract 1 from both sides, but didn't know you could use the Qh, multiply both sides by it to move it from one side of the equation to the other. after finals, i'll go on youtube, and watch all the special algebra movies on Mathtv, the guy explains most of the rules i missed out on learning

cheers AM and that other fella
 

What is a Carnot Engine?

A Carnot Engine is a theoretical heat engine that operates on the Carnot cycle, a reversible thermodynamic cycle. It is used as a model to understand the maximum possible efficiency of an engine.

How does a Carnot Engine work?

A Carnot Engine works by taking in heat from a high temperature reservoir, converting some of it into work, and then releasing the remaining heat into a low temperature reservoir. This process is repeated in a cycle, allowing for continuous work to be done.

What is the efficiency of a Carnot Engine?

The efficiency of a Carnot Engine is given by the equation efficiency = (Thigh - Tlow) / Thigh, where Thigh is the temperature of the high temperature reservoir and Tlow is the temperature of the low temperature reservoir. This means that the efficiency of a Carnot Engine is dependent on the temperature difference between the two reservoirs.

How does a Carnot Engine differ from other heat engines?

A Carnot Engine differs from other heat engines in that it is a theoretical model that operates on a reversible thermodynamic cycle. Other heat engines, such as steam engines or internal combustion engines, operate on irreversible cycles and are not as efficient as a Carnot Engine.

What are some real-world applications of the Carnot Engine?

Although a Carnot Engine is a theoretical model, its principles are used in real-world applications such as refrigerators, air conditioners, and some power plants. These systems use the Carnot cycle to transfer heat from a cooler environment to a warmer one, allowing for cooling or work to be done.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
927
  • Introductory Physics Homework Help
Replies
1
Views
727
  • Introductory Physics Homework Help
Replies
3
Views
873
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
1
Views
721
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
Back
Top