| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
19 Jul 2007 02:05:11 PM |
| Object: |
Physics Question? |
A friend of mine (who doesn't speak good English) gave me this
question. Can anyone answer it? Not sure if the question is phrased
properly:
Question:
The energy stored in a 53.0 HF capacitor is used to melt a 4.00 mg
sample of lead. To what voltage must the capacitor be initally
charged. assuming that the initial temperature of the lead is 20.0 C?
I know that lead has a specific heat of 128 j/kg oC, a melting point
of 327 30C and a latent heat fusion of 24.5 kj/kg.
Thanks for any info.
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| User: "Greg Neill" |
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| Title: Re: Physics Question? |
19 Jul 2007 05:46:51 PM |
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<budgetcomputer@gmail.com> wrote in message
news:1184871911.299103.243430@k79g2000hse.googlegroups.com...
A friend of mine (who doesn't speak good English) gave me this
question. Can anyone answer it? Not sure if the question is phrased
properly:
Question:
The energy stored in a 53.0 HF capacitor is used to melt a 4.00 mg
sample of lead. To what voltage must the capacitor be initally
charged. assuming that the initial temperature of the lead is 20.0 C?
I know that lead has a specific heat of 128 j/kg oC, a melting point
of 327 30C and a latent heat fusion of 24.5 kj/kg.
Thanks for any info.
The energy stored in a capacitor is given by
E = (1/2)*C*V^2
where:
C = capacitance in Farads
V = electric potential in Volts
If you can work out the total energy required to
melt your lead sample (typical chemistry practice
drill), then you can find the required voltage
using the above relationship.
Does anyone have a spare 53 HectoFarad capacitor
I can borrow? :-)
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| User: "Sanny" |
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| Title: Re: Physics Question? |
20 Jul 2007 08:08:09 AM |
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The energy stored in a 53.0 HF capacitor is used to melt a 4.00 mg
sample of lead. To what voltage must the capacitor be initally
charged. assuming that the initial temperature of the lead is 20.0 C?
I know that lead has a specific heat of 128 j/kg oC, a melting point
of 327 30C and a latent heat fusion of 24.5 kj/kg.
Very simple Question just equate the two.
1/2 Cv^2 == Spheat *(meltinpoint-Initial_Temperature ) + latent heat
of fusion.
So V^2 = 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat
of fusion]/C
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2)
Where as you give
Spheat=128 J/Kg`c
meltinpoint=327`c
Initial_Temperature=20`c
latent heat of fusion=24.5 kj/kg = 24500 J/Kg
C=53.0 HF = 5300 F
Putting Values we get
V= { 2*[128 *(327-20) + 24500]/5300 }^(1/2)
Or V={ 2*[128 *(307) + 24500]/5300 }^(1/2)
Or V={ 2*[39296 + 24500]/5300 }^(1/2)
Or V={ 2*[63796]/5300 }^(1/2)
Or V={ 127592/5300 }^(1/2)
Or V={ 24.0739 }^(1/2)
Or V=4.906 ~ 5 Volts.
If the equations are right Answer is 5 Volts.
Bye
Sanny
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| User: "Sanny" |
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| Title: Re: Physics Question? |
20 Jul 2007 08:22:13 AM |
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On Jul 20, 6:08 pm, Sanny <softta...@hotmail.com> wrote:
The energy stored in a 53.0 HF capacitor is used to melt a 4.00 mg
sample of lead. To what voltage must the capacitor be initally
charged. assuming that the initial temperature of the lead is 20.0 C?
I know that lead has a specific heat of 128 j/kg oC, a melting point
of 327 30C and a latent heat fusion of 24.5 kj/kg.
Very simple Question just equate the two.
1/2 Cv^2 == Spheat *(meltinpoint-Initial_Temperature ) + latent heat
of fusion.
So V^2 = 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat
of fusion]/C
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2)
Where as you give
Spheat=128 J/Kg`c
meltinpoint=327`c
Initial_Temperature=20`c
latent heat of fusion=24.5 kj/kg = 24500 J/Kg
C=53.0 HF = 5300 F
Putting Values we get
V= { 2*[128 *(327-20) + 24500]/5300 }^(1/2)
Or V={ 2*[128 *(307) + 24500]/5300 }^(1/2)
Or V={ 2*[39296 + 24500]/5300 }^(1/2)
Or V={ 2*[63796]/5300 }^(1/2)
Or V={ 127592/5300 }^(1/2)
Or V={ 24.0739 }^(1/2)
Or V=4.906 ~ 5 Volts.
If the equations are right Answer is 5 Volts.
Bye
Sanny
One improvement HERE i considered for 1 Kg of Mass. But you want for
4.00 mg of Mass, So the equation need to be modified.
So V^2 = 2*[Spheat *(meltinpoint-Initial_Temperature )*mass + latent
heat of fusion*mass]/C
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature )*mass + latent heat
of fusion*mass]/C }^(1/2)
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2) * [(mass)^(1/2)]
Where as you give
Spheat=128 J/Kg`c
meltinpoint=327`c
Initial_Temperature=20`c
latent heat of fusion=24.5 kj/kg = 24500 J/Kg
C=53.0 HF = 5300 F
Mass= 4 mg = 0.004 Kg
Our Old value was 5 Volt
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2) = 5 Volts.
So just multiply that by [(mass)^(1/2)]
Or V =4.906 * [0.004^(1/2)]
Or V = 4.906 * (0.06324555)
Or V=0.310
So only 0.31 Volts will be sufficient for 4 mg lead. While it will
need 5 Volts for 1 Kg lead.
Bye
Sanny
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| User: "Greg Neill" |
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| Title: Re: Physics Question? |
20 Jul 2007 08:58:53 AM |
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"Sanny" <softtanks@hotmail.com> wrote in message
news:1184937733.355571.151900@z28g2000prd.googlegroups.com...
Our Old value was 5 Volt
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2) = 5 Volts.
So just multiply that by [(mass)^(1/2)]
Or V =4.906 * [0.004^(1/2)]
Careful; 4.00mg = 4.00 x 10^-6 kg, not 4.00 x 10^-3 kg.
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| User: "Sanny" |
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| Title: Re: Physics Question? |
20 Jul 2007 11:20:58 AM |
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I made one mistake 4.00 mg = 4* 10^-6, While I Calculated for 4 gm
Lead by making it 0.004 Kg.
One improvement HERE i considered for 1 Kg of Mass. But you want for
4.00 mg of Mass, So the equation need to be modified.
For 1 Kg it is 5 Volts
For 1 gm it is 0.31 Volts
For 1 mg Lets Calculate.
So V^2 = 2*[Spheat *(meltinpoint-Initial_Temperature )*mass + latent
heat of fusion*mass]/C
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature )*mass + latent heat
of fusion*mass]/C }^(1/2)
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2) * [(mass)^(1/2)]
Where as you give
Spheat=128 J/Kg`c
meltinpoint=327`c
Initial_Temperature=20`c
latent heat of fusion=24.5 kj/kg = 24500 J/Kg
C=53.0 HF = 5300 F
Mass= 4 mg = 0.000004 Kg
Our Old value was 5 Volt for 1 Kg.
V ={ 2*[Spheat *(meltinpoint-Initial_Temperature ) + latent heat of
fusion]/C }^(1/2) = 5 Volts.
So just multiply that by [(mass)^(1/2)]
Or V =4.906 * [0.000004^(1/2)]
Or V = 4.906 * (0.002)
Or V=0.009812 ~ 0.01 Volts
So only 0.01 Volts will be sufficient for melting 4 mg lead, and 0.31
Volts will be sufficient for 4 gm lead. While it will need 5 Volts for
1 Kg lead.
Here we are assuming a few things.
1. No heat loss in Current. That will be given by IR^2
2. No energy loss in energy conversion (100% energy conversion) That
you have to see the manual and find how much heat loss will be there.
Bye
Sanny
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| User: "Eric Gisse" |
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| Title: Re: Physics Question? |
19 Jul 2007 05:38:34 PM |
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On Jul 19, 11:05 am, wrote:
A friend of mine (who doesn't speak good English) gave me this
question. Can anyone answer it? Not sure if the question is phrased
properly:
Question:
The energy stored in a 53.0 HF capacitor is used to melt a 4.00 mg
sample of lead. To what voltage must the capacitor be initally
charged. assuming that the initial temperature of the lead is 20.0 C?
I know that lead has a specific heat of 128 j/kg oC, a melting point
of 327 30C and a latent heat fusion of 24.5 kj/kg.
Thanks for any info.
Your 'friend' gave you a homework problem.
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| User: "H. Wabnig .... .-- .- -... -. .. --. @ .- --- -. DOT .- -" |
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| Title: Re: Physics Question? |
19 Jul 2007 03:08:55 PM |
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On Thu, 19 Jul 2007 12:05:11 -0700, wrote:
A friend of mine (who doesn't speak good English) gave me this
question. Can anyone answer it? Not sure if the question is phrased
properly:
Question:
The energy stored in a 53.0 HF capacitor is used to melt a 4.00 mg
sample of lead. To what voltage must the capacitor be initally
charged. assuming that the initial temperature of the lead is 20.0 C?
I know that lead has a specific heat of 128 j/kg oC, a melting point
of 327 30C and a latent heat fusion of 24.5 kj/kg.
Thanks for any info.
The energy content of a charged cap is
Q = C * U, units are: [Coulomb] or [Ampereseconds]
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