Respuesta :

Explanation:

In each problem, use the linear or volumetric thermal expansion equation.

Linear thermal expansion:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Volumetric thermal expansion:

ΔV = β V₀ ΔT

where ΔL is the change in volume,

β is the volumetric thermal expansion coefficient (β = 3α),

V₀ is the original volume,

and ΔT is the change in temperature.

"What is the length of the same railroad track on a cold winter day when  the temperature is 0°F? "

I assume the railroad track is made of steel, and is originally 30 m at a temperature of 0°C.  Let's first find the change in temperature by converting Fahrenheit to Celsius.

T = 5/9 (0 − 32)

T = -17.8°C

Now we can find the change in length:

ΔL = (11×10⁻⁶ /°C) (30 m) (-17.8 °C − 0°C)

ΔL = -0.006 m

The track shrinks by 0.006 m, so the final length is 24.994 m.

"Estimate the fractional change in the volume of Earth's oceans due to an average temperature change of 1°C."

ΔV = β V₀ ΔT

ΔV / V₀ = β ΔT

ΔV / V₀ = (2.0×10⁻⁴ /°C) (1 °C)

ΔV / V₀ = 2.0×10⁻⁴

"Use the fact that the average depth of the ocean is 4.00×10³ m to estimate the change in depth."

ΔL = α L₀ ΔT

ΔL = (β/3) L₀ ΔT

ΔL = (2.0×10⁻⁴ /°C /3) (4.00×10³ m) ( 1 °C)

ΔL = 0.27 m

"A cylindrical brass sleeve is to be shrink fitted over a brass shaft whose diameter is 3.212 cm at 0°C.  The diameter of the sleeve is 3.196 cm at 0°C.  To what temperature must the sleeve be heated before it will slip over the shaft?"

ΔL = α L₀ ΔT

(3.212 cm − 3.196 cm) = (1.9×10⁻⁵ /°C) (3.196 cm) (T  − 0°C)

T = 263 °C

"Alternatively, to what temperature must the shaft be cooled before it will slip into the sleeve?"

ΔL = α L₀ ΔT

(3.196 cm − 3.212 cm) = (1.9×10⁻⁵ /°C) (3.212 cm) (T  − 0°C)

T = -262 °C

The length of the train track when the temperature is 0°F is 24.994m. One must understand the effects of extreme temperatures on steel. This effect is called Linear Thermal Expansion.

What is the Calculation for the length of the train track when the temperature is 0°?

Linear thermal expansion refers to variations in the dimensions of a solid due to fluctuations in temperature. The formula for Linear Thermal Expansion is given as:

ΔL = α L₀ ΔT: given that

  • ΔL is the change in length of the tracks
  • α is the linear thermal expansion coefficient,
  • L₀ is the original length, and
  • ΔT is the variation in temperature.

Step 1 - Make assumptions

Let's assume that the following:

  • That the track is built of steel
  • That its initial length is 30m;

Step 2 - Insert the figures.

the change in length due to thermal expansion gave as follows:

ΔL = (11×10⁻⁶ /°C) (30 m) (-17.8 °C − 0°C).

ΔL =  -0.006 m

It is to be noted that the temperature was converted from Fahrenheit to Celsius.

Since the original length was 30 and it shrunk by -0.006 m, the new length is 30 - 0.006 = 24.994 m.


What is the Estimates fractional change in the volume of Earth's oceans due to an average temperature change of 1°C.?

This problem is resolved using the knowledge of Volumetric thermal expansion.

What is Volumetric thermal expansion?

The increase in the volume of a solid or liquid when heated is referred to as volumetric thermal expansion.

The formula is given as:

ΔV = β V₀ ΔT

where:

  • ΔL is the change in volume,
  • β is the volumetric thermal expansion coefficient (β = 3α),
  • V₀ is the original volume, and
  • ΔT is the change in temperature.

ΔV = β V₀ ΔT

ΔV / V₀ = β ΔT

ΔV / V₀ = (2.0×10⁻⁴ /°C) (1 °C) Thus

ΔV / V₀ = 2.0×10⁻⁴.


Since the average depth of the ocean is 4.00×10³ m, what is the estimated change in depth?

Step 1 -  Depth is a linear factor. Hence we use the formula

ΔL = α L₀ ΔT

Step 2 - Plug in values

ΔL = (β/3) L₀ ΔT

ΔL = (2.0×10⁻⁴ /°C /3) (4.00×10³ m) ( 1 °C)

Therefore, Estimated Change in Depth is

ΔL = 0.27 m.


To what temperature must the sleeve be heated before it will slip over the shaft?

This also involves Linear Thermal Expansion. Hence,

ΔL = α L₀ ΔT

(3.212 cm − 3.196 cm) = (1.9×10⁻⁵ /°C) (3.196 cm) (T  − 0°C)

T = 263 °C

To what temperature must the shaft be cooled before it will slip into the sleeve?

Recall that the formula is ΔL = α L₀ ΔT.

(3.196 cm − 3.212 cm) = (1.9×10⁻⁵ /°C) (3.212 cm) (T  − 0°C)

T = -262 °C.

Learn more about Linear Thermal Expansion at:
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