| Topic: |
Science > Physics |
| User: |
"V - Man" |
| Date: |
22 Apr 2006 10:16:22 AM |
| Object: |
Physics help |
Hello all. Was wondering if this was the proper forum for getting some help
with a couple of physics questions. The course is physics 201 and the book
is "College Physics" 7th edit. Serway/Vaughn.
If not, sorry for the intrusion.
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| User: "Phil Holman piholmanc@yourservice" |
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| Title: Re: Physics help |
22 Apr 2006 10:25:05 AM |
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Most folks here are OK with providing help as long as you show your
attempt at the problem.
"V - Man" <Vman@nowhere.com> wrote in message
news:e2dhbv$8b0d$1@news3.infoave.net...
Hello all. Was wondering if this was the proper forum for getting some
help with a couple of physics questions. The course is physics 201 and
the book is "College Physics" 7th edit. Serway/Vaughn.
If not, sorry for the intrusion.
Most folks here are OK with providing help as long as you show your
attempt at the problem.
Phil H
.
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| User: "V - Man" |
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| Title: Re: Physics help |
22 Apr 2006 10:42:47 AM |
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Most folks here are OK with providing help as long as you show your
attempt at the problem.
Thanks for replying, Phil. Unfortunately, my attempt is futile. I have no
idea how to approach solving this problem. Our prof. is a chemistry prof.
and has never taught physics before. So, I am trying with no success to
answer this one question based on his limited lecture notes. The text is not
much of a help either. All we get are the formulas with no problems to
practice putting the formulas to work!!
Here's the problem (if anyone can help, great! If not, I understand):
"I take 1.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium
is reached, I have 2.0 kg of ice at 0 deg C. The water was originally at 0
deg C. The specific heat of water = 1.00 kcal/kg*deg C, the specific heat of
ice = 0.50 kcal/kg*C, and the latent heat of fusion of water is 80 kcal/kg.
The original temp of the ice was.....??????"
If anyone would just maybe be kind enough to even help setting it up, I
would really appreciate it. Thanks and again, sorry for the inrusion.
.
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| User: "John Christiansen" |
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| Title: Re: Physics help |
22 Apr 2006 11:12:38 AM |
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"V - Man" <Vman@nowhere.com> skrev i en meddelelse
news:e2ditp$8cp7$1@news3.infoave.net...
Most folks here are OK with providing help as long as you show your
attempt at the problem.
Thanks for replying, Phil. Unfortunately, my attempt is futile. I have no
idea how to approach solving this problem. Our prof. is a chemistry prof.
and has never taught physics before. So, I am trying with no success to
answer this one question based on his limited lecture notes. The text is
not much of a help either. All we get are the formulas with no problems to
practice putting the formulas to work!!
Here's the problem (if anyone can help, great! If not, I understand):
"I take 1.0 kg of ice and dump it into 1.0 kg of water and, when
equilibrium is reached, I have 2.0 kg of ice at 0 deg C. The water was
originally at 0 deg C. The specific heat of water = 1.00 kcal/kg*deg C,
the specific heat of ice = 0.50 kcal/kg*C, and the latent heat of fusion
of water is 80 kcal/kg. The original temp of the ice was.....??????"
If anyone would just maybe be kind enough to even help setting it up, I
would really appreciate it. Thanks and again, sorry for the inrusion.
An important thing to notice is that the temperature of the water does not
change. All that happens is that the water is transformed into ice, and it
happens by giving off energy, and that energy is used to increase the
temperature of the ice. It is all that happens.
Energy released from water is mass of water multiplied by latent heat of
fusion
Rise in temperature of ice is available energy divided by specific heat of
ice divided by mass of ice
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| User: "V - Man" |
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| Title: Re: Physics help |
22 Apr 2006 11:59:37 AM |
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"> An important thing to notice is that the temperature of the water does
not
change. All that happens is that the water is transformed into ice, and it
happens by giving off energy, and that energy is used to increase the
temperature of the ice. It is all that happens.
Energy released from water is mass of water multiplied by latent heat of
fusion
Rise in temperature of ice is available energy divided by specific heat of
ice divided by mass of ice
Thanks, John. I guess I have a tendancy to overthink the problem. Since the
water stayed the same temperature, just had a phase change, the ice must
have been only 1 or 2 degrees below C.
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| User: "Greg Neill" |
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| Title: Re: Physics help |
22 Apr 2006 12:41:08 PM |
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"V - Man" <Vman@nowhere.com> wrote in message news:e2dndh$8hr6$1@news3.infoave.net...
"> An important thing to notice is that the temperature of the water does
not
change. All that happens is that the water is transformed into ice, and it
happens by giving off energy, and that energy is used to increase the
temperature of the ice. It is all that happens.
Energy released from water is mass of water multiplied by latent heat of
fusion
Rise in temperature of ice is available energy divided by specific heat of
ice divided by mass of ice
Thanks, John. I guess I have a tendancy to overthink the problem. Since the
water stayed the same temperature, just had a phase change, the ice must
have been only 1 or 2 degrees below C.
Latent heat of fusion for water is 80kcal/kg. So you need
to dump 1kg x 80kcal/kg = 80kcals into the ice. You should
be able to work out the temperature change in the 1kg of
ice that corresponds to this amount of heat energy.
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| User: "Timo Nieminen" |
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| Title: Re: Physics help |
22 Apr 2006 04:36:50 PM |
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On Sat, 22 Apr 2006, V - Man wrote:
"> An important thing to notice is that the temperature of the water does
not
change. All that happens is that the water is transformed into ice, and it
happens by giving off energy, and that energy is used to increase the
temperature of the ice. It is all that happens.
Energy released from water is mass of water multiplied by latent heat of
fusion
Rise in temperature of ice is available energy divided by specific heat of
ice divided by mass of ice
Thanks, John. I guess I have a tendancy to overthink the problem. Since the
water stayed the same temperature, just had a phase change, the ice must
have been only 1 or 2 degrees below C.
No, there's a lot of energy involved in phase changes. Do the numbers and
see. Sometimes a guess of "only 1 or 2 degrees" is wrong.
(I thought specific heat capacity, latent heat etc were taught in
chemistry as well as physics.)
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "V - Man" |
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| Title: Re: Physics help |
22 Apr 2006 04:41:30 PM |
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No, there's a lot of energy involved in phase changes. Do the numbers and
see. Sometimes a guess of "only 1 or 2 degrees" is wrong.
(I thought specific heat capacity, latent heat etc were taught in
chemistry as well as physics.)
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Well, I now know that 1 or 2 degrees is wrong. However, after I work the
problem, I end up with -160 degrees C. It seems too cold. Not sure if I have
inserted to correct numbers.
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| User: "Timo Nieminen" |
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| Title: Re: Physics help |
22 Apr 2006 05:59:05 PM |
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On Sat, 22 Apr 2006, V - Man wrote:
No, there's a lot of energy involved in phase changes. Do the numbers and
see. Sometimes a guess of "only 1 or 2 degrees" is wrong.
Well, I now know that 1 or 2 degrees is wrong. However, after I work the
problem, I end up with -160 degrees C. It seems too cold. Not sure if I have
inserted to correct numbers.
The numbers seem reasonable (c for ice is correct, and Lf should be
about 100 times larger, so appears OK). If I fill a glass with ice cubes
colder than 0 degrees C, then top up with water, I don't see the water
freeze. Try an experiment, and see for yourself. It isn't easy to get 1 kg
of ice at -160C, so do it in reverse, using hot water to melt ice. Get 1
kg of ice cubes (that will give you a larger surface area than a single 1
kg block; 1 kg of crushed ice might be even better). The heat capacity of
water is about double that of ice, so get 1 litre (ie 1 kg) of water at
about 80C, and it should melt the ice, leaving you with very cold water.
Try it. Boiling water should leave you with water at about 10C, colder
than 80C should leave you with ice water. Of course, it'll warm up from
the surroundings as well as from the hot water, but you'll get an idea of
the amount of energy required to melt ice.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "V - Man" |
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| Title: Re: Physics help |
22 Apr 2006 06:33:38 PM |
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The numbers seem reasonable (c for ice is correct, and Lf should be
about 100 times larger, so appears OK). If I fill a glass with ice cubes
colder than 0 degrees C, then top up with water, I don't see the water
freeze. Try an experiment, and see for yourself. It isn't easy to get 1 kg
of ice at -160C, so do it in reverse, using hot water to melt ice. Get 1
kg of ice cubes (that will give you a larger surface area than a single 1
kg block; 1 kg of crushed ice might be even better). The heat capacity of
water is about double that of ice, so get 1 litre (ie 1 kg) of water at
about 80C, and it should melt the ice, leaving you with very cold water.
Try it. Boiling water should leave you with water at about 10C, colder
than 80C should leave you with ice water. Of course, it'll warm up from
the surroundings as well as from the hot water, but you'll get an idea of
the amount of energy required to melt ice.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Timo, thank you so much for responding and also, adding to further help me
understand about phase changes. It is all very fascinating for sure. When I
originally got the -160 C I thought for sure I made a mistake. However, I
now realize the significance of how much energy is needed for phase changes.
Pretty amazing!
.
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| User: "Henning Makholm" |
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| Title: Re: Physics help |
22 Apr 2006 05:50:24 PM |
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Scripsit "V - Man" <Vman@nowhere.com>
Since the water stayed the same temperature, just had a phase
change, the ice must have been only 1 or 2 degrees below C.
The word "just" in this sentence is unsupported, and represents a fallacy.
--
Henning Makholm "I ... I have to return some videos."
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| User: "Gregory L. Hansen" |
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| Title: Re: Physics help |
22 Apr 2006 01:12:37 PM |
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In article <e2ditp$8cp7$1@news3.infoave.net>, V - Man <Vman@nowhere.com> wrote:
Most folks here are OK with providing help as long as you show your
attempt at the problem.
Thanks for replying, Phil. Unfortunately, my attempt is futile. I have no
idea how to approach solving this problem. Our prof. is a chemistry prof.
and has never taught physics before. So, I am trying with no success to
answer this one question based on his limited lecture notes. The text is not
much of a help either. All we get are the formulas with no problems to
practice putting the formulas to work!!
Here's the problem (if anyone can help, great! If not, I understand):
"I take 1.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium
is reached, I have 2.0 kg of ice at 0 deg C. The water was originally at 0
deg C. The specific heat of water = 1.00 kcal/kg*deg C, the specific heat of
ice = 0.50 kcal/kg*C, and the latent heat of fusion of water is 80 kcal/kg.
The original temp of the ice was.....??????"
If anyone would just maybe be kind enough to even help setting it up, I
would really appreciate it. Thanks and again, sorry for the inrusion.
As John Christiansen noted, the temperature of the water does not change.
Another thing to note is that none of the original chunk of ice melts.
The heat given up by the water is
Q_water = L m_water
where L is the heat of fusion and m is the mass.
The heat absorbed by the ice is
Q_ice = C m_ice (T_f - T_i)
where T_i is the initial temperature, and T_f = 0C is the final
temperature.
Q_water = Q_ice, so, after sorting out the sign convention, solving the
problem becomes simple algebra.
--
"He who only sees business in business is a fool."
.
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| User: "V - Man" |
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| Title: Re: Physics help |
22 Apr 2006 01:55:39 PM |
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As John Christiansen noted, the temperature of the water does not change.
Another thing to note is that none of the original chunk of ice melts.
The heat given up by the water is
Q_water = L m_water
where L is the heat of fusion and m is the mass.
The heat absorbed by the ice is
Q_ice = C m_ice (T_f - T_i)
where T_i is the initial temperature, and T_f = 0C is the final
temperature.
Q_water = Q_ice, so, after sorting out the sign convention, solving the
problem becomes simple algebra.
--
"He who only sees business in business is a fool."
So, working out the problem I get Q_water = 80
For Q-ice: C m_ice = 0.50 * 1 (0 - T_i)
After working out the simple algebra, I got -160 C. This doesn't seem
logical. Unless the C m_ice is not correct. Should the m be the final amount
of ice or starting block? If final (2 kg), then the answer is -80 C.
.
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| User: "Henning Makholm" |
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| Title: Re: Physics help |
22 Apr 2006 05:25:19 PM |
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Scripsit "V - Man" <Vman@nowhere.com>
After working out the simple algebra, I got -160 C. This doesn't seem
logical.
Why not? Freezing the water will release *a lot* of heat; if you can
pump *all* of it into the original ice without warming it more than to
0 °C, it must have been pretty cold to start with.
In fact, the didactic point of the problem may be to impress you with
how much more energy is involved in a phase change than in everyday
temperature variation within a phase.
--
Henning Makholm "What the hedgehog sang is not evidence."
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| User: "V - Man" |
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| Title: Re: Physics help |
22 Apr 2006 04:32:51 PM |
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"I take 1.0 kg of ice and dump it into 1.0 kg of water and, when
equilibrium
is reached, I have 2.0 kg of ice at 0 deg C. The water was originally at 0
deg C. The specific heat of water = 1.00 kcal/kg*deg C, the specific heat
of
ice = 0.50 kcal/kg*C, and the latent heat of fusion of water is 80
kcal/kg.
The original temp of the ice was.....??????"
The heat given up by the water is
Q_water = L m_water
where L is the heat of fusion and m is the mass.
The heat absorbed by the ice is
Q_ice = C m_ice (T_f - T_i)
where T_i is the initial temperature, and T_f = 0C is the final
temperature.
Is the "C" in the Q_ice formula = to 0.5????? And I assume the m is 1 kg.
I can't imagine the final answer is -160 C for the starting temp of the ice!
Q_water = Q_ice, so, after sorting out the sign convention, solving the
problem becomes simple algebra.
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