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?