I never argued that heat isn't a form of energy, or that it isn't
Well, Bob argues that it's power that heats up a resistor not energy..."Kelvin@!!!" <kelvin604@shaw.ca.ca> wrote in message
news:JFgnc.421393$Ig.277708@pd7tw2no...
"Bob Penoyer" <rpenoyer@NOSPAMieee.org> wrote in message
news:m1qq90t6dqecfj69c0f99g6435ih0pi0vp@4ax.com...
On Sat, 08 May 2004 05:41:40 GMT, "Kelvin@!!!"
kelvin604@shaw.ca.ca
wrote:
"Bob Penoyer" <rpenoyer@NOSPAMieee.org> wrote in message
Well, let's see. If I dump 1000 joules into a 1-watt resistor
but I
take all of February to do it, the resistor won't get very hot.
But,
if I do it over a period of 100 seconds, the resistor will get
damn
hot. Rate! Rate is important. It's power that heats the
resistor.
Hence, terms like 1-WATT resistor.
i wonder if you did your lil dumping-joul experiment in VACUMM
AND there
is
no heat RADIATING from that resistor!!!
Heat radiates just fine from a resistor in a vacuum. If heat
couldn't
radiate through a vacuum, the sun wouldn't do us much good.
in vacuum and radiation are different statements... they are not a
bit
related...
joul is the unit of energy, also the unti of heat!!!
which means... no matter how many jouls you dump into a object...
as long as there's no heat lost... the temp of that object will
go up by
a
certain degrees...
A resistor won't be insulated under normal circumstances. So heat
will
be lost by conduction, convection, and/or radiation.
for the case you were talking about... it's becuase a partial
heat that
CURRENT generated is LOST in air and radiation... that's why you
dont
feel
it really hot if you take a whole month to dump that 1000J into
the
resistor...
for only 100s... you feel it much hotter becuase not much heat is
running
away~~~
Nonsense. As I said in my earlier post, it is rate that effects
the
heating of the resistor. If you dump energy into the resistor
faster
(i.e., apply more power), the resistor will get hotter.
alrite... one simple Q
do u measure heat in Jouls or in Watts?
have a nice day~~~
You, too.
--
{ Kelvin@!!! }
Stop it, you are making my head hurt...
The amount of power radiated from a resistor is related to its
temperature. The higher the temp, in relation to the environment, the
faster the object will radiate energy.
If you add energy by passing current through it, it will heat up until
the rate of radiation = the rate of addition, ie, it will reach a
thermal equilibrium.
If it is unable to radiate energy for some reason (maybe the
surroundings are kept at the same increasing temperature) the resistor
will continue to heat up, because energy is getting added by the
current flow.
Thus, you are both right, it's energy that heats a resistor, but in
the real world, its power that determines the final temperature,
because, given a particular resistor configuration, the resistor will
radiate at a particular rate dependent on temperature. Thus, the
temperature will satisfy a formula something like
P(T) = I^2*R
By 'radiation' I was talking about all forms of energy loss. For this--
{ Kelvin@!!! }
"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:Y3jnc.9065$iF6.1024231@attbi_s02...
"Kelvin@!!!" <kelvin604@shaw.ca.ca> wrote in message
news:JFgnc.421393$Ig.277708@pd7tw2no...
"Bob Penoyer" <rpenoyer@NOSPAMieee.org> wrote in message
news:m1qq90t6dqecfj69c0f99g6435ih0pi0vp@4ax.com...
On Sat, 08 May 2004 05:41:40 GMT, "Kelvin@!!!"
kelvin604@shaw.ca.ca
wrote:
"Bob Penoyer" <rpenoyer@NOSPAMieee.org> wrote in message
Well, let's see. If I dump 1000 joules into a 1-watt
resistor
but I
take all of February to do it, the resistor won't get very
hot.
But,
if I do it over a period of 100 seconds, the resistor will
get
damn
hot. Rate! Rate is important. It's power that heats the
resistor.
Hence, terms like 1-WATT resistor.
i wonder if you did your lil dumping-joul experiment in
VACUMM
AND there
is
no heat RADIATING from that resistor!!!
Heat radiates just fine from a resistor in a vacuum. If heat
couldn't
radiate through a vacuum, the sun wouldn't do us much good.
in vacuum and radiation are different statements... they are not
a
bit
related...
joul is the unit of energy, also the unti of heat!!!
which means... no matter how many jouls you dump into a
object...
as long as there's no heat lost... the temp of that object
will
go up by
a
certain degrees...
A resistor won't be insulated under normal circumstances. So
heat
will
be lost by conduction, convection, and/or radiation.
for the case you were talking about... it's becuase a partial
heat that
CURRENT generated is LOST in air and radiation... that's why
you
dont
feel
it really hot if you take a whole month to dump that 1000J
into
the
resistor...
for only 100s... you feel it much hotter becuase not much
heat is
running
away~~~
Nonsense. As I said in my earlier post, it is rate that
effects
the
heating of the resistor. If you dump energy into the resistor
faster
(i.e., apply more power), the resistor will get hotter.
alrite... one simple Q
do u measure heat in Jouls or in Watts?
You, too.
--
{ Kelvin@!!! }
Stop it, you are making my head hurt...
The amount of power radiated from a resistor is related to its
temperature. The higher the temp, in relation to the environment,
the
faster the object will radiate energy.
If you add energy by passing current through it, it will heat up
until
the rate of radiation = the rate of addition, ie, it will reach a
thermal equilibrium.
If it is unable to radiate energy for some reason (maybe the
surroundings are kept at the same increasing temperature) the
resistor
will continue to heat up, because energy is getting added by the
current flow.
Thus, you are both right, it's energy that heats a resistor, but
in
the real world, its power that determines the final temperature,
because, given a particular resistor configuration, the resistor
will
radiate at a particular rate dependent on temperature. Thus, the
temperature will satisfy a formula something like
Well, Bob argues that it's power that heats up a resistor not
energy...
and one more point in your arguement...
heat is loosing by convention, conduction, and radiation...
radiation is not the only way that heat is transfered from one body
to
another...
the "T" in my equation above is clearly stated to be temperature. P(T)I^2*R = P(T)
where P(T) is the function of radiated power for the object in its
environment, given a temperature T.
Regards,
Bob Monsen
P(T) = I^2*R
where is the independent variable T(time) in the RHS ??? so time is
not
involved in this equation...
I think you mean
P(I) = I^2*R
or
W(T) = P*T
hv a nice day~~~
No you guys, you have T for both time and temperature!! Silly!--
{ Kelvin@!!! }
"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:Y3jnc.9065$iF6.1024231@attbi_s02...
"Kelvin@!!!" <kelvin604@shaw.ca.ca> wrote in message
news:JFgnc.421393$Ig.277708@pd7tw2no...
"Bob Penoyer" <rpenoyer@NOSPAMieee.org> wrote in message
news:m1qq90t6dqecfj69c0f99g6435ih0pi0vp@4ax.com...
On Sat, 08 May 2004 05:41:40 GMT, "Kelvin@!!!"
kelvin604@shaw.ca.ca
wrote:
"Bob Penoyer" <rpenoyer@NOSPAMieee.org> wrote in message
Well, let's see. If I dump 1000 joules into a 1-watt resistor
but I
take all of February to do it, the resistor won't get very hot.
But,
if I do it over a period of 100 seconds, the resistor will get
damn
hot. Rate! Rate is important. It's power that heats the
resistor.
Hence, terms like 1-WATT resistor.
i wonder if you did your lil dumping-joul experiment in VACUMM
AND there
is
no heat RADIATING from that resistor!!!
Heat radiates just fine from a resistor in a vacuum. If heat
couldn't
radiate through a vacuum, the sun wouldn't do us much good.
in vacuum and radiation are different statements... they are not a
bit
related...
joul is the unit of energy, also the unti of heat!!!
which means... no matter how many jouls you dump into a object...
as long as there's no heat lost... the temp of that object will
go up by
a
certain degrees...
A resistor won't be insulated under normal circumstances. So heat
will
be lost by conduction, convection, and/or radiation.
for the case you were talking about... it's becuase a partial
heat that
CURRENT generated is LOST in air and radiation... that's why you
dont
feel
it really hot if you take a whole month to dump that 1000J into
the
resistor...
for only 100s... you feel it much hotter becuase not much heat is
running
away~~~
Nonsense. As I said in my earlier post, it is rate that effects
the
heating of the resistor. If you dump energy into the resistor
faster
(i.e., apply more power), the resistor will get hotter.
alrite... one simple Q
do u measure heat in Jouls or in Watts?
You, too.
--
{ Kelvin@!!! }
Stop it, you are making my head hurt...
The amount of power radiated from a resistor is related to its
temperature. The higher the temp, in relation to the environment, the
faster the object will radiate energy.
If you add energy by passing current through it, it will heat up until
the rate of radiation = the rate of addition, ie, it will reach a
thermal equilibrium.
If it is unable to radiate energy for some reason (maybe the
surroundings are kept at the same increasing temperature) the resistor
will continue to heat up, because energy is getting added by the
current flow.
Thus, you are both right, it's energy that heats a resistor, but in
the real world, its power that determines the final temperature,
because, given a particular resistor configuration, the resistor will
radiate at a particular rate dependent on temperature. Thus, the
temperature will satisfy a formula something like
Well, Bob argues that it's power that heats up a resistor not energy...
and one more point in your arguement...
heat is loosing by convention, conduction, and radiation...
radiation is not the only way that heat is transfered from one body to
another...
I^2*R = P(T)
where P(T) is the function of radiated power for the object in its
environment, given a temperature T.
Regards,
Bob Monsen
P(T) = I^2*R
where is the independent variable T(time) in the RHS ??? so time is not
involved in this equation...
I think you mean
* P(I) = I^2*R
or
W(T) = P*T
hv a nice day~~~
-------------------
Heat radiates from the earth out into open space with
Not true. You can (and do) radiate into a surrounding medium
Well, true, but not salient.
is arguingsnip
That should have read, " No. Only by the U. S. Army.""John Popelish" <jpopelish@rica.net> wrote in message
news:409AC096.F7F74067@rica.net...
JeffM wrote:
John,
Have you ever been paid for teaching by an institution of (higher?)
learning?
Only by the U. S. Army.
John, I am SO tempted at this point, but I'm going to let
this one pass by...it's just too much of an easy target...