Question

A 110-kg steel support rod in a building has a length of 2.0 m at a temperature of 20°C. The rod supports a hanging load of 5,000 kg. (a) Find the work done on the rod as the temperature increases to 37°C. (Assume the average coefficient of linear expansion for steel is 11 x 10-6 (°C) -7.) (b) Find the energy Q added to the rod (assume the specific heat of steel is the same as that for iron, 448 J/(kg • °C) and enter your answer to at least six significant figures). (c) Find the change in internal energy of the rod. (Enter your answer to at least six significant figures). Need Help? Read It

Answer 1

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PART (a):

Given that,

Length of the rod, L = 2 m

Coefficient of linear expansion, $\alpha=11\times10^{-6}\;(^{o}C)^{-1}$

Change in temperature, $\Delta T=37^{o}C-20^{o}C=17^{o}C$

Therefore, Change in length of the rod is given by:

$\Delta L=L.\alpha.\Delta T=2\;m\times 11\times10^{-6}\;(^{o}C)^{-1}\times17^{o}C$

$\therefore \Delta L=3.74\times10^{-4}\;m$

Also, force is given by:

$F=mg=5000\;kg\times9.8\;m/s^{2}=49000\;N$

Therefore, Work done on the rod is given by:

$W=F.\Delta L=49000\;N\times3.74\times10^{-4}\;m$

$\boldsymbol{\therefore W=18.326\;J}$

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PART (b):

Given that,

Mass of the rod, M = 110 kg.

Specific heat capacity, C = 448 J/kg.^oC.

Therefore, Energy added to the rod is given by:

$Q=MC\Delta T=110\;kg\times448\;J/kg^{o}C\times17^{o}C$

$\boldsymbol{\therefore Q=837760\;J}$

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PART (c):

Change in internal energy of the rod is given by:

$\Delta U=Q-W=837760\;J-18.326\;J$

$\boldsymbol{\therefore \Delta U=837741.674\;J}$

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