Dolphin 發問時間： 社會與文化語言 · 1 0 年前

# 請問有關物理浮力的相關問題? 急！急！急！緊急！

1. An inventor constructs a thermometer from an aluminum bar that is

0.500m in length at 273k.He measures the temperature by measuring the length of the aluminum bar. If the inventor wants to measure a 1k change in temperature, how precisely must he measure the length of the bar?

2. Bridge builders often use rivets that are larger than the rivet hole to make the joint righter. The rivet is cooled before it is put into the hole. Suppose that a builder drills a hole 1.2230 cm in diameter for a steel rivet

1.2250 cm in diameter.To what temperature must the rivet be cooled if it is to fit into the rivet hole, which is at 20.0 度 c?

3. A square of a copper is heated from 11度c to 580度 c. If the volume of the sphere is 1.78cm^3 at 11度c, what is the increase in volume of the

sphere at 580度c?

### 1 個解答

• 1 0 年前
最佳解答

This is about themal expansion not buoyant force.

1. Linear thermal expansion coefficient of aluminum is 23.1 &thinsp;µm/mK at 25 degree Celsius. It is 0.000 0231 meter per meter per degree Celsius. The aluminum bar is 0.5000m in original. For 1 degree higher it will expand 0.000 01155 meter long. He must measure as precisely as 0.000 001 meter, it means the instrument for measuring is readable to 0.000 001 meter. However, the original length is only measured to 0.500m not 0.500 000m. therefore, it is nonsense to make any measurement preciser than the original length.

2. Linear thermal expansion coefficient of iron and steel is 12 &thinsp;µm/mK at 20 degree Celsius. Assumed this coeffiicent valid down to 137 degrees Celsius

1.2250cm-1.2230cm= 0.0020cm

0.0020cm/1.2250cm = 0.001632653cm/cm

(0.001632653 cm/cm)/(12 &thinsp;µm/mK) = 136.054 K

Need to cold down 136.054 degrees Celsius and a little bit more for clearance. In addition, need to make sure the thermal expansion coefficient at that low temperature.

3. Linear thermal expansion coefficient of copper is at 20 degree Celsius. Assumed this coeffiicent valid through 20 to 580 Celsius

Volime of sphere is 4(pi)(r^3) = 4*3.14*r*r*r

the original volume = 1.78cm^3, so, original r = 0.52127cm

after heated to 580 Celsius, r= 0.52127cm*(580K-11K)*17&thinsp;µm/mK = 0.526321112cm

Volime of sphere at 580 Celsius is 4*3.14*0.526321112^3cm^3 = 1.83215cm^3 = 1.83cm^3

volume increased 0.05215cm^3

2007-02-14 17:42:01 補充：

2. The rivet shall be colded down 136K from existing 20 degree Celsius, namely, 20 - 136 = - 116, colded down to a temperature of - 116 degree Celsius.

2007-02-14 17:44:55 補充：

please show out all of your answers. then I will check them all.

2007-02-14 18:21:12 補充：

the 3rd one should to corrected to be:

V= (4/3) pi r^3

if V = 1.78cm^3, then original r = 0.751813cm

from 11 degree C to 580 degree C, the r increases to be 0.75*(1 plus 569*0.000017) = 0.759086cm, this is heated radius.

2007-02-14 18:21:36 補充：

(continued)

Then the heated volume is (4/3)*pi*r^3 = (4/3)*3.1416* 0.759086cm^3 = 1.832155cm^3

The volume increases 1.83 - 1.78 = 0.05cm^3.

2007-02-14 18:40:38 補充：

1.

If the coefficient of thermal expansion is 0.0000231 per degree Celsius then the increment is 0.0000231 * 0.500m = 0.00001165m, therefore the measuring tape should be readable to 0.000 001m.

1K = 0.000 011 55 m

2K = 0.000 023 10 m

3K = 0.000 034 65 m

4K = 0.000 046 20 m