a cylindrical lead rod witha diameter of 2.54 cm and a length of 76.2 cm is init

Question

a cylindrical lead rod witha diameter of 2.54 cm and a length of 76.2 cm is initially at a temperature of 20.0 degrees celcius. Next 2.27 kg of mercury at a temperature of 357 degrees celcuis is poured over the rod.

a) after reaching eqilipbrium, what is the final temperature of teh rod and the merrcury?

b) after reaching eqilibrium, what is the length of the rod in cm?

c) after reaching equilibrium, what is the volume of the rod?

to get points please please show detailed equations and steps used!

Explanation / Answer

Part A)

By conservation of energy, the heat lost by the mercury is gained by the lead rod

We need the mass of the lead using the density

density = m/V

density of lead is 11.3 X 10^3 kg/m^3

mass = (11.3 X 10^3)(pi)(.0127)^2(.762)

mass = 4.363 kg

mc(delta T) = mc(delta T)

(4.363)(128)(Tf - 20) = (2.27)(138)(357 - Tf)

558.5Tf - 11169.3 = 111834 - 313.3Tf

871.8Tf = 123003.3

Tf = 141 degrees C

Part B)

Delta L = Lo(alpha)(delta T)

Delta L = (.762)(29 X 10^-6)(141 - 20)

Delta L = 2.67 X 10^-3 m which is 2.67 mm

Part C)

Delta V = Vo(Beta)(delta T)

Delta V = (pi)(.0127^2)(.762)(3)(29 X 10^-6)(141-20)

Delta V = 4.06 X 10^-6 m^3 which is 4.06 cm^3

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