wire has a diameter of o.03589 inches. Solid rhodium has a density pont o gens\'

Question

wire has a diameter of o.03589 inches. Solid rhodium has a density pont o gens' and licluid rhodium has a density of 10 650 gem, at 1966.the meting of 12.410 point of thodium. The current price ol (20 points) or150.0 t of 19-gauge thodium wire. Recall: Density ons un a. The length of the wire in centimeters b. The diameter of the wire in centimeters The volume of the wire in cubic cent mt alur c The volume of the wire in cubic centimeter, V-arl (05-0826 n d. The weight of the wire in pounds. 1 lb 2205 kg 6.001 ke

Explanation / Answer

Ans. #c. Volume of 150-ft (=4572.0 cm) wire = (3.14159) x (0.09116 cm/ 2)2 x 4572 cm

                                                                        = 29.8405296 cm3

#e. From #c, we have, the volume of solid wire = 29.8405296 cm3

Now,

            Mass of wire = Volume x Density of solid metal

                                    = 29.8405296 cm3 x (12.410 g/ cm3)

                                    = 370.320972336 g

# The mass of metal remains constant irrespective of its physical state (solid or liquid).

So, mass of molten (liquid) metal when wire melts = 370.320972336 g

Now,

            Volume of liquid metal = Mass / Density of liquid metal

                                    = 370.320972336 g / (10.650 g / cm3)

                                    = 34.7719222 cm3

# Change in volume = Volume of liquid metal – Volume of solid metal

                                    = 34.7719222 cm3 – 29.8405296 cm3

                                    = (+) 4.9313926 cm3

Note that the (+) sign indicates that the volume of metal increase when the solid melts into liquid form.

#f. Given, rate of metal wire = $950 / tr Oz = $950 / 31.103 g

Now,

            Mass of desired wire costing $250 = Available cost / Rate of wire

                                                = $250 / ($950 / 31.103 g)

                                                = 8.185 g

            Volume of desired (solid) wire = 8.185 g / (12.410 g/ cm3) = 0.6595 cm3

# We have, diameter of 19-gauge wire = 0.09116 cm

                        Volume of desired wire = 0.6595 cm3

Let the desired length of wire be l cm.

So,

            0.6595 cm3 = (3.14159) x (0.09116 cm/ 2)2 x l

            Or, l = 0.6595 cm3 / (0.0065268 cm2)

            Hence, l = 101.045 cm

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