An 8 000-kg aluminum flagpole 100-m long i heated by the un from a temperature of 10°C to 20°C. Find the work done (in J) by the aluminum if the linear expanion coefficient i 24 ´ 10-6 (°C)-1. (The denity of aluminum i 2. 7 ´ 103 kg/m3 and 1 atm = 1. 0 ´ 105 N/m2



Answer :

The work done by the aluminum in 216 Joules.

The ratio of the increase in length to the original length for each degree increase in temperature is called the coefficient of linear expansion. The coefficient of linear thermal expansion describes the change in the length of a material as a function of temperature. A distinction is made between the average thermal coefficient of expansion and the physical coefficient of linear expansion.

calculation:-

Work done = msT2 -T1

= 8000000 g × 2 × 1/000000 ( 20- 10)

= 600

density = 2.7

work done per unit = 600/2.7

= 216 jolues

The volume expansion coefficient is three times the linear expansion coefficient. Heating an object changes its dimensions. It depends on your body type. Expansion can be done in length in which case it is called linear expansion. Heating a square wafer causes expansion in two planes lengthwise and widthwise. We call this range dilation.

Learn more about The linear expansion coefficient here:-https://brainly.com/question/19495810

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