please I need help on this assignment.
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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.
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
Step 1 - Make assumptions
Let's assume that the following:
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.
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:
ΔV = β V₀ ΔT
ΔV / V₀ = β ΔT
ΔV / V₀ = (2.0×10⁻⁴ /°C) (1 °C) Thus
ΔV / V₀ = 2.0×10⁻⁴.
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.
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
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:
https://brainly.com/question/6985348
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