- #1
jrd007
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I did all the word but my answers are not matching my books?
1) Can a 2.5 mm-diameter copper wire have the same resistence as a tungsten wire of the same length? Give numerical detials. Answer, yes, 4.6 mm
Here is what I did:
( p = resistivity )
R = p(L/A) so I set R(Cu) = R(tug) and since L is the same it cancels out.
Area = 1/2pie2^2 ---> Cu area = (.5)(3.14)(1.25mm) ---> A = .393 mm
So p(A) = p(A) ---> (1.68 x 10^-8)(.393 mm) = (5.6 x 10^-8)(A)
A = .1179 mm
therefore diameter would be... .1179 mm = 2(pie)r^2
r = .2729 mm which mean diameter = .546 mm... so what did I do incorrect?
2) A small immersion heater can be used in a car to heat a cup of water. If the heater can heat 120 mL of water from 25C to 95 C in 8.0 minutes, (a) How much current does it draw from the car's 12 V battery, (b) What is the resistence? Assume the manufacturer's clain of 60% effciency. Answers: 10 A, 1.2 Olhms
What I did...
Used specific heat equation ---> Q = mcT where I changed 120 mL of water into .120 kg ---> Q = (.120 kg)(4185 J/kg C)(95C-25C)
Q = 35154 J
Then I used the equation PE = 1/2Q(where Q is the charger)V
35151 J = (.5)(Q)(12 V)
Q = 5859 C
Finally I used I(current) = Q/T(time) ---> 5859C/480s
I = 12 A
And for Resistence I used V/I = R ---> 12 V/12 A = 1 Olhm
I am off my two units for some odd reason in my current and off by .2 in my resistence. Did I make a mistake?
1) Can a 2.5 mm-diameter copper wire have the same resistence as a tungsten wire of the same length? Give numerical detials. Answer, yes, 4.6 mm
Here is what I did:
( p = resistivity )
R = p(L/A) so I set R(Cu) = R(tug) and since L is the same it cancels out.
Area = 1/2pie2^2 ---> Cu area = (.5)(3.14)(1.25mm) ---> A = .393 mm
So p(A) = p(A) ---> (1.68 x 10^-8)(.393 mm) = (5.6 x 10^-8)(A)
A = .1179 mm
therefore diameter would be... .1179 mm = 2(pie)r^2
r = .2729 mm which mean diameter = .546 mm... so what did I do incorrect?
2) A small immersion heater can be used in a car to heat a cup of water. If the heater can heat 120 mL of water from 25C to 95 C in 8.0 minutes, (a) How much current does it draw from the car's 12 V battery, (b) What is the resistence? Assume the manufacturer's clain of 60% effciency. Answers: 10 A, 1.2 Olhms
What I did...
Used specific heat equation ---> Q = mcT where I changed 120 mL of water into .120 kg ---> Q = (.120 kg)(4185 J/kg C)(95C-25C)
Q = 35154 J
Then I used the equation PE = 1/2Q(where Q is the charger)V
35151 J = (.5)(Q)(12 V)
Q = 5859 C
Finally I used I(current) = Q/T(time) ---> 5859C/480s
I = 12 A
And for Resistence I used V/I = R ---> 12 V/12 A = 1 Olhm
I am off my two units for some odd reason in my current and off by .2 in my resistence. Did I make a mistake?