{"id":1804,"date":"2018-04-11T11:42:20","date_gmt":"2018-04-11T03:42:20","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1804"},"modified":"2023-08-09T08:39:39","modified_gmt":"2023-08-09T08:39:39","slug":"thermal-expansion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/thermal-expansion-problems-and-solutions.htm","title":{"rendered":"Thermal expansion – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Thermal expansion – problems and solutions<\/span><\/p>\n

Area expansion<\/strong><\/a><\/span><\/p>\n

1. A sheet of steel at 20o<\/sup>C has size as shown in the figure below. If the coefficient of l<\/a>inear expansion for steel is 10-5<\/sup> o<\/sup>C-1<\/sup> then what is the change in the area at 60o<\/sup>C.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Length of steel = 40 cm \"Thermal<\/span><\/p>\n

Width of steel = 20 cm <\/span><\/p>\n

The initial of steel’s area (Ao<\/sub>) = (40)(20) = 800 cm2<\/sup><\/span><\/p>\n

The coefficient of linear expansion (\u03b1) = 10-5<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of area expansion (\u03b2) = 2 x coefficient of linear expansion (2\u03b1) = 2 x 10-5<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The change in temperature (\u0394T) = 60o<\/sup>C \u2013 20o<\/sup>C = 40o<\/sup>C<\/span><\/p>\n

Wanted :<\/u> The change in area of steel at 60o<\/sup>C<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Equation of area expansion :<\/span><\/p>\n

\u0394A = \u03b2 Ao <\/sub>\u0394T <\/span><\/p>\n

\u0394A = the increase in area of steel, \u03b2 = The coefficient of area expansion, Ao<\/sub> = initial area, \u0394T = the change in temperature = final temperature \u2013 initial temperature<\/span><\/p>\n

The increase <\/u>in area of steel :<\/u><\/span><\/p>\n

\u0394A = \u03b2 Ao <\/sub>\u0394T <\/span><\/p>\n

\u0394A = (2 x 10-5<\/sup>)(800)(40) = 0.64 cm <\/span><\/p>\n

2. A plate of iron at 20o<\/sup>C has shown in figure below. If the temperature is raised to 100o<\/sup>C and the coefficient of linear expansion of iron is 1.1 x 10-7<\/sup> o<\/sup>C-1<\/sup>, then what is the final area of plate.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Length of plate = 2 m \"Thermal<\/span><\/p>\n

Width of plate = 2 m <\/span><\/p>\n

The initial area of iron (Ao<\/sub>) = (2)(2) = 4 m2<\/sup><\/span><\/p>\n

The coefficient of linear expansion for iron (\u03b1) = 1.1 x 10-7<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of area expansion for iron (\u03b2) = 2 x the coefficient of linear expansion for iron (2\u03b1) = 2.2 x 10-7<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The change in temperature (\u0394T) = 100o<\/sup>C \u2013 20o<\/sup>C = 80o<\/sup>C<\/span><\/p>\n

Wanted <\/u>:<\/u> Area of iron at 100o<\/sup>C<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The increase in length :<\/span><\/p>\n

\u0394A = \u03b2 Ao <\/sub>\u0394T <\/span><\/p>\n

\u0394A = (2.2 x 10-7<\/sup>)(4)(80) = 704 x 10-7 <\/sup>= 0,0000704 m2<\/sup> <\/span><\/p>\n

Area of iron :<\/u><\/span><\/p>\n

Area of iron = initial area + the increase in area<\/span><\/p>\n

Area of iron = 4 m2 <\/sup>+ 0.0000704 m2<\/sup> <\/span><\/p>\n

Area of iron = 4.0000704 m2<\/sup><\/span><\/p>\n

3. A bronze plate with the coefficient of linear expansion \u03b1 = 18.10-6<\/sup> o<\/sup>C-1 <\/sup>at 0o<\/sup>C has size as shown in figure below. If the plate heated at 80 o<\/sup>C, then what is the increase in area of plate.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The length of bronze = 40 cm = 0.4 meters\"Thermal<\/span><\/p>\n

Width of bronze = 20 cm = 0.2 meters<\/span><\/p>\n

Initial area of bronze (Ao<\/sub>) = (0.4)(0.2) = 0.08 m2<\/sup><\/span><\/p>\n

The coefficient of linear expansion for bronze (\u03b1) = 18 x 10-6<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of area expansion for bronze (\u03b2) = 2 x The coefficient of linear expansion (2\u03b1) = 36 x 10-6<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The change in temperature (\u0394T) = 80o<\/sup>C \u2013 0o<\/sup>C = 80o<\/sup>C<\/span><\/p>\n

Wanted :<\/u> The increase of area for bronze at 80o<\/sup>C<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The increase of area for bronze :<\/span><\/p>\n

\u0394A = \u03b2 Ao <\/sub>\u0394T <\/span><\/p>\n

\u0394A = (36 x 10-6<\/sup>)(0.08)(80) = 230.4 x 10-6<\/sup> = 2.304 x 10-4<\/sup> m2<\/sup><\/span><\/p>\n

Volume expansion<\/strong><\/a><\/span><\/p>\n

4. A glass container with volume of 4 liters filled with water, then heated until the increase in temperature is 20o<\/sup>C. Some water spilled. The coefficient of linear expansion for glass = 9 x 10-6<\/sup> o<\/sup>C-1<\/sup>; the coefficient of volume expansion for water = 2.1 x 10-4<\/sup> o<\/sup>C-1<\/sup>. Determine the volume of spilled water.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The initial volume of the gas and water (Vo<\/sub>) = 4 liters<\/span><\/p>\n

The increase in temperature of the glass and water (\u0394T) = 20o<\/sup>C<\/span><\/p>\n

The coefficient of linear expansion for glass (\u03b1) = 9 x 10-6<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of volume expansion for glass (\u03b3) = 3\u03b1 = 3 (9 x 10-6<\/sup> o<\/sup>C-1<\/sup>) = 27 x 10-6<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of volume expansion for water (\u03b3) = 2.1 x 10-4<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

Wanted :<\/u> Volume of spilled water<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of the volume expansion :<\/span><\/p>\n

V = Vo<\/sub> + \u03b3 Vo<\/sub> \u0394T <\/span><\/p>\n

V – Vo<\/sub> = \u03b3 Vo<\/sub> \u0394T <\/span><\/p>\n

\u0394V = \u03b3 Vo<\/sub> \u0394T <\/span><\/p>\n

V = <\/i>final volume<\/i>, <\/i>V<\/i>o<\/i><\/sub> = <\/i>initial volume<\/i>, <\/i>\u0394V = <\/i>the change in volume<\/i>, <\/i>\u03b3 = <\/i>the coefficient of volume expansion<\/i>, <\/i>\u0394T = <\/i>the change in temperature.<\/i><\/span><\/p>\n

The change in volume of the glass container :<\/span><\/p>\n

\u0394V = \u03b3 Vo<\/sub> \u0394T = (27 x 10-6<\/sup>)(4)(20) = 2160 x 10-6<\/sup> = 2.160 x 10-3<\/sup> = 0.002160 liters<\/span><\/p>\n

The change in volume of the water :<\/span><\/p>\n

\u0394V = \u03b3 Vo<\/sub> \u0394T = (2.1 x 10-4<\/sup>)(4)(20) = 168 x 10-4<\/sup> = 0.0168 liters<\/span><\/p>\n

The change in volume of the water is greater than the glass container, so some water spills.<\/i><\/span><\/p>\n

The volume of spilled water :<\/span><\/p>\n

0.0168 liters \u2013 0.002160 liters = 0.01464 liters = 0.015 liters<\/span><\/p>\n

5. A steel container (the coefficient of linear expansion = 10-5<\/sup> o<\/sup>C-1<\/sup>) with volume of 6 liters filled with acetone (the coefficient of volume expansion = 1.5 x 10-3<\/sup> o<\/sup>C-1<\/sup>). If the container and acetone are heated from 0o<\/sup>C to 40o<\/sup>C, what is the volume of spilled acetone?<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The initial volume of the container and acetone (Vo<\/sub>) = 6 liters<\/span><\/p>\n

The change in temperature of the container and acetone (\u0394T) = 40o<\/sup>C<\/span><\/p>\n

The coefficient of linear expansion for steel (\u03b1) = 10-5<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of volume expansion for steel (\u03b3) = 3\u03b1 = 3 (10-5<\/sup> o<\/sup>C-1<\/sup>) = 3 x 10-5<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

The coefficient of volume expansion for acetone (\u03b3) = 1.5 x 10-3<\/sup> o<\/sup>C-1<\/sup><\/span><\/p>\n

Wanted :<\/u> The volume of spilled acetone<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of volume expansion :<\/span><\/p>\n

\u0394V = \u03b3 Vo<\/sub> \u0394T <\/span><\/p>\n

\u0394V = <\/i>the change in volume<\/i>, <\/i>\u03b3 = <\/i>the coefficient of volume expansion<\/i>, V<\/i>o<\/i><\/sub> = <\/i>initial volume<\/i>, <\/i>\u0394T = <\/i>the change in temperature<\/i>.<\/i><\/span><\/p>\n

The change in volume of the steel container :<\/span><\/p>\n

\u0394V = \u03b3 Vo<\/sub> \u0394T = (3 x 10-5<\/sup>)(6)(40) = 720 x 10-5 <\/sup>= 0.00720 liters<\/span><\/p>\n

The change in volume of the acetone :<\/span><\/p>\n

\u0394V = \u03b3 Vo<\/sub> \u0394T = (1.5 x 10-3<\/sup>)(6)(40) = 360 x 10-3<\/sup> = 0.360 liters<\/span><\/p>\n

The change in volume of the <\/i>acetone <\/i>is greater than the <\/i>steel <\/i>container, so some <\/i>acetone s<\/i>pills.<\/i><\/span><\/p>\n

The volume of acetone spilled :<\/span><\/p>\n

0.360 liters \u2013 0.00720 liters = 0.3528 liters = 0.35 liters<\/span><\/p>\n

    \n
  1. What is thermal expansion?<\/strong><\/span>\n