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Physics Help

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Finns14

Golden Member
Ok so you guys have been great in the past so I'm hoping for some more help now here is the first problem

The hot and cold reservoirs of a Carnot engine have temperatures of 1010 and 291 K, respectively. The engine does the work of lifting a 21.1-kg block straight up from rest, so that at a height of 15.6 m the block has a speed of 7.47 m/s. How much heat must be put into the engine?

The answer is 5358.292308692629 J

So can anyone point me in the right direction? Thanks in advanced!
 
Figure out how much work it requires to lift the block from rest up to the 15.6 m height with the 7.47 m/s velocity. Then use the temperatures you're given and the effeciency of a Carnot engine to determine how much heat you need to put in to get the work required to lift the block out.
 
Yeah I was leading in this direction then I was wondering do you set the work calculated over the unknown?
 
Originally posted by: Finns14
Yeah I was leading in this direction then I was wondering do you set the work calculated over the unknown?
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I'm not sure what you mean. Just do (work required to lift the block)=(Carnot efficiency)*(heat needed to be put into engine).
 
Exactly. You have to use the Heisenberg Pinciple.

Sorry, the coeficient of friction was not high enough to stop me from saying that in a physics thread 😛
 
Hmm I seem to be getting stuck some where I am geting close but off. I calcualte eff first by adding the 2 temps together and then taking the first temp and dividing it by the sum of the temps. Then to calculate work I and taking the distance and dividing it by the rate to find accel. then multipling it by the mass and distance. Then the equation is eff= 1- (work/x) where am I going wrong?
 
Originally posted by: Evadman
Exactly. You have to use the Heisenberg Pinciple.

Sorry, the coeficient of friction was not high enough to stop me from saying that in a physics thread 😛
Click to expand...
If you're not careful I'll whack you upside the head with the coefficient of friction. 😛
 
calculate the work needed to lift object to 15.6 meters high (use Force x distance)
calculate work needed to give object velocity of 7.47 m/s (use kinetic energy equation)

equate the sum of the work to efficiency of the cycle times x (x being the energy put in)

I worked it out and the answer is off by single digits
 
Originally posted by: Finns14
Hmm I seem to be getting stuck some where I am geting close but off. I calcualte eff first by adding the 2 temps together and then taking the first temp and dividing it by the sum of the temps. Then to calculate work I and taking the distance and dividing it by the rate to find accel. then multipling it by the mass and distance. Then the equation is eff= 1- (work/x) where am I going wrong?
Click to expand...
Well you can use this page to check your efficiency number. As for the work value, I'm not going to work it out, but I think you've got the general idea of what to do to get the answer. 🙂
 
Still not getting it where am I going wrong I think its calculating the work, can anyone be more specific?
 
Originally posted by: Heisenberg
Originally posted by: Finns14
Hmm I seem to be getting stuck some where I am geting close but off. I calcualte eff first by adding the 2 temps together and then taking the first temp and dividing it by the sum of the temps. Then to calculate work I and taking the distance and dividing it by the rate to find accel. then multipling it by the mass and distance. Then the equation is eff= 1- (work/x) where am I going wrong?
Click to expand...
Well you can use this page to check your efficiency number. As for the work value, I'm not going to work it out, but I think you've got the general idea of what to do to get the answer. 🙂
Click to expand...

Its saying I am calculating my eff wrong?

Even with the "correct" eff I am still off.
 
work due to lifting = 21.1 kg * 9.81 m/s^2 * 15.6 m = 3229.0596 J
work due to velocity/kinetic energy = 0.5 * 21.1 kg * (7.47 m/s)^2 = 588.699495 J

efficiency of cycle = 1 - (291/1010) = 0.71188

heat added * efficiency = 3229.0596 J + 588.699495 J

head added = (3229.0596 J + 588.699495 J) / 0.71188 = approximately to the answer above

you may want to use a more specific value of gravity

The problem is so theoretical that the precision of the answer should not even matter. Whoever expects a theoretical answer with that much precision is a nut-case.
 
Originally posted by: Atomicus
work due to lifting = 21.1 kg * 9.81 m/s^2 * 15.6 m = 3229.0596 J
work due to velocity/kinetic energy = 0.5 * 21.1 kg * (7.47 m/s)^2 = 588.699495 J

efficiency of cycle = 1 - (291/1010) = 0.71188

heat added * efficiency = 3229.0596 J + 588.699495 J

head added = (3229.0596 J + 588.699495 J) / 0.71188 = approximately to the answer above

you may want to use a more specific value of gravity
Click to expand...

Yep, very good answer.
 
Originally posted by: Atomicus
work due to lifting = 21.1 kg * 9.81 m/s^2 * 15.6 m = 3229.0596 J
work due to velocity/kinetic energy = 0.5 * 21.1 kg * (7.47 m/s)^2 = 588.699495 J

efficiency of cycle = 1 - (291/1010) = 0.71188

heat added * efficiency = 3229.0596 J + 588.699495 J

head added = (3229.0596 J + 588.699495 J) / 0.71188 = approximately to the answer above

you may want to use a more specific value of gravity
Click to expand...



Ok I was an idiot and forgot A was grav but how are you calcualting eff cause the site provide is giving me a diff number


Edit once again I am a dumbass about the eff
 
Originally posted by: Finns14
AHHH this is sooo confusing
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I'm looking through this semester's notes in Power Systems and there is not other explanation in your variation in efficiency. Even the site that Heisenberg posted gives the same efficiency as doing 1 - (291/1010). The Carnot cycle is very idealistic, which is why the efficiency formula is so generic.

Like I said before, Whoever expects a theoretical answer with that much precision is a nut-case.
 
3.43 kilograms of liquid water at 0 °C is put into the freezer compartment of a Carnot refrigerator. The temperature of the compartment is -17.2 °C, and the temperature of the kitchen is 17.2 °C. If the cost of electrical energy is 0.150 dollars per kilowatt-hour, how many dollars does it cost to make 3.43 kilograms of ice at 0 °C?
 
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