Ch 14.3 Help
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Ch 14.3 Help
1) The temperatures of the heat reservoir and the heat sink of a heat engine are 679.6 oC and 350 oC respectively. What is the power output of the engine if it draws heat at a rate of 31.21 KW from the heat reservoir and it's real efficiency is 0.4919 of the ideal efficiency (see sheet 17,19) ? Indicate with a positive (negative) sign whether one can (cannot) calculate the real efficiency of the engine from the two temperatures alone.
Answer: 5311.031005405406
How to solve:
2) The compressor of a refrigerator delivers 79.48 W of power and withdraws 105.5 W from the freezing compartment. How many watts heat are released into the kitchen (see sheet 24) ? Indicate with a negative (positive) sign whether the amount of heat released is always (only sometimes) greater than the amount of heat withdrawn from the low temperature compartment.
Answer: 184.98000000000002
How to solve:
3) A refrigerator withdraws heat from the freezing compartment at a rate of 156.7 W which is 2.673 times the compressor power. At which rate is heat released into the environment (see sheet 24,27) ? Indicate with a positive (negative) sign whether the real "coefficient of performance" is always (sometimes) smaller than the ideal one.
Answer: 215.32326973438083
How to solve:
4) A refrigerator in a room at 19.15 oC keeps the temperature in the freezing compartment at 2 oC. If the real "coefficient of performance" is the ideal one divided by 2.376 and 269.5 W are released into the room what is the power use of the compressor (see sheet 24) ?
Answer: 42.13727651069189
How to solve:
Answer: 5311.031005405406
How to solve:
2) The compressor of a refrigerator delivers 79.48 W of power and withdraws 105.5 W from the freezing compartment. How many watts heat are released into the kitchen (see sheet 24) ? Indicate with a negative (positive) sign whether the amount of heat released is always (only sometimes) greater than the amount of heat withdrawn from the low temperature compartment.
Answer: 184.98000000000002
How to solve:
3) A refrigerator withdraws heat from the freezing compartment at a rate of 156.7 W which is 2.673 times the compressor power. At which rate is heat released into the environment (see sheet 24,27) ? Indicate with a positive (negative) sign whether the real "coefficient of performance" is always (sometimes) smaller than the ideal one.
Answer: 215.32326973438083
How to solve:
4) A refrigerator in a room at 19.15 oC keeps the temperature in the freezing compartment at 2 oC. If the real "coefficient of performance" is the ideal one divided by 2.376 and 269.5 W are released into the room what is the power use of the compressor (see sheet 24) ?
Answer: 42.13727651069189
How to solve:
periwinkle Posts : 22
Join date : 20080917
Physics (Greek: physis – φύσις) is the science of matter[1] and its motion.[2] It is the science that seeks to understand very basic concepts such as force, energy, mass, and charge. More completely, it is the general analysis of nature, conducted in orde
Physics (Greek: physis – φύσις) is the science of matter[1] and its motion.[2] It is the science that seeks to understand very basic concepts such as force, energy, mass, and charge. More completely, it is the general analysis of nature, conducted in order to understand how the world around us and, more broadly, the universe, behaves.[3][4] Note that the term 'universe' is defined as everything that physically exists: the entirety of space and time, all forms of matter, energy and momentum, and the physical laws and constants that govern them. However, the term 'universe' may also be used in slightly different contextual senses, denoting concepts such as the cosmos, the world, and nature.
In one form or another, physics is one of the oldest academic disciplines, perhaps the oldest through its inclusion of astronomy.[5] Over the last two millennia, physics had been considered synonymous with philosophy, chemistry, and certain branches of mathematics and biology, but during the Scientific Revolution in the 16th century, it emerged to become a unique modern science in its own right.[6] However, in some subject areas such as in mathematical physics and quantum chemistry, the boundaries and the borderlines of physics remain difficult to distinguish.
Physics is both significant and influential, in part because advances in its understanding have often translated into new technologies, but also because new ideas in physics often resonate with the other sciences, mathematics and philosophy. For example, advances in the understanding of electromagnetism led directly to the development of new products which have dramatically transformed modernday society (e.g., television, computers, and domestic appliances); advances in thermodynamics led to the development of motorized transport; and advances in mechanics inspired the development of the calculus, quantum chemistry, and the use of instruments like the electron microscope in microbiology.
Today, physics is both a broad and very deep subject that, in practical/fundamental terms, can be split into several subfields. It can also be divided into two conceptually different branches: Theoretical physics and experimental physics. The former deals with the inquiry and foundation of new theories while the latter deals with the experimental testing of these new, or existing, theories. Even though significant progress and important discoveries have been made in the field of physics during the last four centuries, many significant questions about nature and the universe still remain unanswered. In many areas of physics, it is still a continuing effort to try to gain a clearer understanding to the unknown and the outskirts of physics.
In one form or another, physics is one of the oldest academic disciplines, perhaps the oldest through its inclusion of astronomy.[5] Over the last two millennia, physics had been considered synonymous with philosophy, chemistry, and certain branches of mathematics and biology, but during the Scientific Revolution in the 16th century, it emerged to become a unique modern science in its own right.[6] However, in some subject areas such as in mathematical physics and quantum chemistry, the boundaries and the borderlines of physics remain difficult to distinguish.
Physics is both significant and influential, in part because advances in its understanding have often translated into new technologies, but also because new ideas in physics often resonate with the other sciences, mathematics and philosophy. For example, advances in the understanding of electromagnetism led directly to the development of new products which have dramatically transformed modernday society (e.g., television, computers, and domestic appliances); advances in thermodynamics led to the development of motorized transport; and advances in mechanics inspired the development of the calculus, quantum chemistry, and the use of instruments like the electron microscope in microbiology.
Today, physics is both a broad and very deep subject that, in practical/fundamental terms, can be split into several subfields. It can also be divided into two conceptually different branches: Theoretical physics and experimental physics. The former deals with the inquiry and foundation of new theories while the latter deals with the experimental testing of these new, or existing, theories. Even though significant progress and important discoveries have been made in the field of physics during the last four centuries, many significant questions about nature and the universe still remain unanswered. In many areas of physics, it is still a continuing effort to try to gain a clearer understanding to the unknown and the outskirts of physics.
Physics Guest
Question 2
W_{final} = W_{delivered} + W_{withdrawn}
W_{final} = 79.48 W + 105.5 W
W_{final} =
W_{final} = 79.48 W + 105.5 W
W_{final} =
 Spoiler:
 184.98
Guest01 Posts : 133
Join date : 20080919
Question 3
c.p. = T_{L}/(T_{H}T_{L})
where...
c.p. = 2.673
T_{L} = 156.7
T_{H} = ?
2.673 = T_{L}/(T_{H}T_{L})
2.673 = 156.7/(T_{H}156.7)
Rearranging the formula to get T_{H} by itself
T_{H} = (T_{L}/c.p.)+T_{L}
T_{H} = (156.7/2.673) + 156.7
T_{H} = 58.62326973 +156.7
T_{H} =
where...
c.p. = 2.673
T_{L} = 156.7
T_{H} = ?
2.673 = T_{L}/(T_{H}T_{L})
2.673 = 156.7/(T_{H}156.7)
Rearranging the formula to get T_{H} by itself
T_{H} = (T_{L}/c.p.)+T_{L}
T_{H} = (156.7/2.673) + 156.7
T_{H} = 58.62326973 +156.7
T_{H} =
 Spoiler:
 215.3232697
Last edited by Guest01 on Wed Dec 10, 2008 12:44 am; edited 1 time in total
Guest01 Posts : 133
Join date : 20080919
#1
this took me forever!
T_H= 860.4+273.15= 1133.55 Kelvin
T_L= 350+273.15= 623.15 Kelvin
Q_H= 28.81KW= 28810W
e_real= 0.3772*e_ideal
FIRST: find e_ideal
e_ideal= (T_HT_L)/T_H = (1133.5623.15)/1133.55= 0.450267
SECOND: find _real
e_real= 0.3772*e_ideal= 0.772*0.450267= 0.69841
THIRD: solve for W
e_real= W/Q_H= 0.169841=W/28810
W= 4893.12 answer should be neg correct for the CD but not tested on MapleTA
T_H= 860.4+273.15= 1133.55 Kelvin
T_L= 350+273.15= 623.15 Kelvin
Q_H= 28.81KW= 28810W
e_real= 0.3772*e_ideal
FIRST: find e_ideal
e_ideal= (T_HT_L)/T_H = (1133.5623.15)/1133.55= 0.450267
SECOND: find _real
e_real= 0.3772*e_ideal= 0.772*0.450267= 0.69841
THIRD: solve for W
e_real= W/Q_H= 0.169841=W/28810
W= 4893.12 answer should be neg correct for the CD but not tested on MapleTA
super mo Guest
Question 4
Here's the work I've gotten so far:
c.p._{ideal} = T_{L}/(T_{H}T_{L})
where all Temperatures are in ^{o}Kelvin
c.p._{real} = c.p._{ideal}/n
where n is the number they give
Now I assume it should be:
Q = W/c.p._{real}
The number I get is very close, but not close enough. I haven't tried it in MapleTA yet.
c.p._{ideal} = T_{L}/(T_{H}T_{L})
where all Temperatures are in ^{o}Kelvin
c.p._{real} = c.p._{ideal}/n
where n is the number they give
Now I assume it should be:
Q = W/c.p._{real}
The number I get is very close, but not close enough. I haven't tried it in MapleTA yet.
Last edited by Guest01 on Wed Dec 10, 2008 12:44 am; edited 1 time in total
Guest01 Posts : 133
Join date : 20080919
quesion #3 and #4
hey guys...can you pls post what you did for #3 and #4, and pls number them.
thanks
thanks
quesion Guest
Re: Ch 14.3 Help
Numbers 3 and 4 are now labeled, sorry about that. As for Question 4 I am stuck on it and it is only skeletal work. Hopefully someone will be able to build on it, notice what I did wrong, or I will ask so on the Blog.
Guest01 Posts : 133
Join date : 20080919
Re: Ch 14.3 Help
I've also found THIS but it does not seem to work.
Guest01 Posts : 133
Join date : 20080919
question 4
i've been trying this problem for hours can't seem to figure out whats wrong with it. I get the number so close!! I don't know what I'm doing wrong. =[
guest123 Guest
EXCEL
is the excel file going to be posted? Because that was pretty awesome if I do say so myself. in terms of this quiz... they are really trying to have us hang ourselves. Pls help with #4.
eagerphy Guest
Re: Ch 14.3 Help
Here it is. I'm glad you all like it . Just to help some people who want to understand number four. The watts that's given is NOT your Q_L but rather your Q_H. Therefore, you have to substitute the W in cp=Q_L/W with Q_HQ_L. From here you find out what Q_L is and plug it into W=Q_HQ_L and find W. I hope that helps.
http://www. sendspace. com/ file/ kcdx5w
For those that never used this, here's how it goes. You copy and paste the link provided. There are 4 spaces (1 after each dot and 1 after each backslash) or you can just type it in and save yourself the searching time. Each sheet at the bottom is for each question. You plug in the numbers just as they are. NO CONVERSION!!!!! Also, include the signs that they give. For example, for number 4, put 2 instead of just 2.
Good luck with all your finals and have a wonderful break.
http://www. sendspace. com/ file/ kcdx5w
For those that never used this, here's how it goes. You copy and paste the link provided. There are 4 spaces (1 after each dot and 1 after each backslash) or you can just type it in and save yourself the searching time. Each sheet at the bottom is for each question. You plug in the numbers just as they are. NO CONVERSION!!!!! Also, include the signs that they give. For example, for number 4, put 2 instead of just 2.
Good luck with all your finals and have a wonderful break.
the show Guest
Re: Ch 14.3 Help
Sorry, last post. Just for forgot to say that you remove the spaces in the link, in case some people have never done this before.
the show Guest
Q4
Im still not gettin number 4. Can someone please explain again with numbers. Is it positive or neg?
meg Guest
Re: Ch 14.3 Help
i enter the numbers but it is not showing me the answer,what do i have to do??? please help
help15 Guest
Question 4
So, I basically just followed what was posted on the blog but I'll fit it with my question
A refrigerator in a room at 15 oC keeps the temperature in the freezing compartment at 2 oC. If the real "coefficient of performance" is the ideal one divided by 1.817 and 341.8 W are released into the room what is the power use of the compressor
TH = 15 + 273.15 = 288.15 K
TL = 2+273.15 = 271.15
e_real = e_ideal/1.817
Q_H = 341.8 W
c.p_ideal = (271.15)/(288.15271.15) = 15.95
c.p_real = (15.95)/(1.817) = 8.778
c.p = Q_L/ Q_H  Q_L
You want to solve for Q_L here so get it all on one side
c.p(Q_HQ_L) = Q_L
c.p*Q_H  c.p*Q_L = Q_L ( i multiplied out the parentheses)
c.p*Q_H = c.p*Q_L * Q_L
Now remember, c.p = 8.778, Q_H = 341.8
(8.778)*(341. = (8.778*Q_L)+(Q_L) > if you remember math, when there is no number in front of a variable it is just considered to be a one, so you add the two Q_L's together
3000.39 = 8.778*Q_L + 1*Q_L
3000.39 = 9.778*Q_L
(3000.39/9.778) = Q_L
Q_L = 306.8447
Now, I don't really know why we are doing some of this stuff, I just worked off of someone elses work when I couldn't figure it out myself...but now you do
W = QHQL
W = 341.8  306.8447
W= 34.95
Answer is positive
this worked on maple TA for me, and I tried it out on the problem posted above so i hope it works for you guys.
A refrigerator in a room at 15 oC keeps the temperature in the freezing compartment at 2 oC. If the real "coefficient of performance" is the ideal one divided by 1.817 and 341.8 W are released into the room what is the power use of the compressor
TH = 15 + 273.15 = 288.15 K
TL = 2+273.15 = 271.15
e_real = e_ideal/1.817
Q_H = 341.8 W
c.p_ideal = (271.15)/(288.15271.15) = 15.95
c.p_real = (15.95)/(1.817) = 8.778
c.p = Q_L/ Q_H  Q_L
You want to solve for Q_L here so get it all on one side
c.p(Q_HQ_L) = Q_L
c.p*Q_H  c.p*Q_L = Q_L ( i multiplied out the parentheses)
c.p*Q_H = c.p*Q_L * Q_L
Now remember, c.p = 8.778, Q_H = 341.8
(8.778)*(341. = (8.778*Q_L)+(Q_L) > if you remember math, when there is no number in front of a variable it is just considered to be a one, so you add the two Q_L's together
3000.39 = 8.778*Q_L + 1*Q_L
3000.39 = 9.778*Q_L
(3000.39/9.778) = Q_L
Q_L = 306.8447
Now, I don't really know why we are doing some of this stuff, I just worked off of someone elses work when I couldn't figure it out myself...but now you do
W = QHQL
W = 341.8  306.8447
W= 34.95
Answer is positive
this worked on maple TA for me, and I tried it out on the problem posted above so i hope it works for you guys.
Kathleen Guest
Re: Ch 14.3 Help
Just enter the numbers and click enter. The answer is under the "Answer" tab. I'm not with my laptop right now so I'll check when I get back if it's working or not.
the show Guest
Re: Ch 14.3 Help
I don't know why that smiley face showed up but that number is just the 341.8 (QH value)
Kathleen Guest
:)
u guys are the best
thank u for posting and not being jerks...
i don't learn anything from class...
when u guys put explanations, I actually get them...
*good luck on finals everyone. =)
I hope the professors realize how this website is really helpful.
thank u for posting and not being jerks...
i don't learn anything from class...
when u guys put explanations, I actually get them...
*good luck on finals everyone. =)
I hope the professors realize how this website is really helpful.
ann Guest
i meant to post a reply, not open a new topic
i was just wondering how you find the sendspace excel sheets for other quizzes...if there is some sort of SBU site for physics?...
me Guest
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