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Thermal Expansion of Solids (Linear, Area, and Volume Expansivity)

16 questions

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Question 1Question →

A solid brass cube with an edge length of 10 cm10\text{ cm} at 15∘C15^\circ\text{C} is heated to a temperature of 115∘C115^\circ\text{C}. If the linear expansivity of brass is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}, what is the increase in the volume of the cube in cm3\text{cm}^3?

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Answer: 6

Answer

The increase in the volume of the brass cube is 6.0 cm36.0\text{ cm}^3.
The volume expansion of a solid is given by ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T. The initial volume of the cube is V1=(10 cm)3=1000 cm3V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3 and the temperature change is ΔT=115∘C−15∘C=100 K\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}. Because the expansion occurs in three dimensions, the volume expansivity is γ=3α=3×2.0×10−5=6.0×10−5 K−1\gamma = 3\alpha = 3 \times 2.0 \times 10^{-5} = 6.0 \times 10^{-5}\text{ K}^{-1}. Substituting these values yields ΔV=1000×6.0×10−5×100=6.0 cm3\Delta V = 1000 \times 6.0 \times 10^{-5} \times 100 = 6.0\text{ cm}^3.

Step-by-Step Solution

1
Calculate the initial volume of the cube
V1=(10 cm)3=1000 cm3V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3
The volume of a cube is calculated using V=L3V = L^3, where LL is the edge length.
2
Determine the change in temperature
ΔT=115∘C−15∘C=100 K\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}
The change in temperature is the difference between the final and initial temperatures.
3
Calculate the volume expansivity (cubic expansivity)
γ=3α=3×(2.0×10−5 K−1)=6.0×10−5 K−1\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}
Volume expansivity γ\gamma is three times the linear expansivity α\alpha for isotropic solids.
4
Compute the increase in volume
ΔV=V1γΔT=1000×(6.0×10−5)×100=6.0 cm3\Delta V = V_1 \gamma \Delta T = 1000 \times (6.0 \times 10^{-5}) \times 100 = 6.0\text{ cm}^3
The formula for volume expansion is ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T.

Key Concept

Thermal expansion of solids: volume expansion (ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T) and the relationship between linear and volume expansivity (γ=3α\gamma = 3\alpha).
Question 2Question →

An iron ring has an internal cross-sectional area of 0.50 m20.50\text{ m}^2 at 30∘C30^\circ\text{C}. If the linear expansivity of iron is 1.2×10−5 K−11.2 \times 10^{-5}\text{ K}^{-1}, what is the increase in its internal cross-sectional area when heated to 130∘C130^\circ\text{C}?

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Answer: 1.2×10−3 m21.2 \times 10^{-3}\text{ m}^2

Answer

The increase in the internal cross-sectional area of the ring is 1.2×10−3 m21.2 \times 10^{-3}\text{ m}^2.
For two-dimensional (area) expansion of solids, the area expansivity β\beta is equal to twice the linear expansivity (2α2\alpha). Given α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}, β=2.4×10−5 K−1\beta = 2.4 \times 10^{-5}\text{ K}^{-1}. Multiplying by the initial area (0.50 m20.50\text{ m}^2) and temperature rise (100 K100\text{ K}) yields an area increase of 1.2×10−3 m21.2 \times 10^{-3}\text{ m}^2.

Step-by-Step Solution

1
Determine the temperature change (ΔT\Delta T) and the superficial expansivity (β\beta).
ΔT=130∘C−30∘C=100 K\Delta T = 130^\circ\text{C} - 30^\circ\text{C} = 100\text{ K}, and β=2α=2×(1.2×10−5 K−1)=2.4×10−5 K−1\beta = 2\alpha = 2 \times (1.2 \times 10^{-5}\text{ K}^{-1}) = 2.4 \times 10^{-5}\text{ K}^{-1}.
Area expansion depends on superficial expansivity, which is twice the linear expansivity for an isotropic solid.
2
Calculate the increase in area (ΔA\Delta A) using the area expansion formula.
ΔA=A0βΔT=0.50 m2×(2.4×10−5 K−1)×100 K=1.2×10−3 m2\Delta A = A_0 \beta \Delta T = 0.50\text{ m}^2 \times (2.4 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 1.2 \times 10^{-3}\text{ m}^2.
The fractional change in area is directly proportional to initial area, superficial expansivity, and temperature change.

Key Concept

Relationship between linear expansivity (α\alpha) and superficial expansivity (β=2α\beta = 2\alpha) in thermal expansion of area.
Estimated Time:1m 30s
Question 3Question →

A metal rod has a linear expansivity of 1.5×10−5 K−11.5 \times 10^{-5}\text{ K}^{-1}. What is the volume expansivity of a sphere made from the same metal?

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Answer: 4.5×10−5 K−14.5 \times 10^{-5}\text{ K}^{-1}

Answer

The volume expansivity of the sphere is 4.5×10−5 K−14.5 \times 10^{-5}\text{ K}^{-1}.
The volume expansivity γ\gamma of a uniform solid object is related to its linear expansivity α\alpha by γ=3α\gamma = 3\alpha. Multiplying 1.5×10−5 K−11.5 \times 10^{-5}\text{ K}^{-1} by 33 gives 4.5×10−5 K−14.5 \times 10^{-5}\text{ K}^{-1}.

Step-by-Step Solution

1
Identify the mathematical relationship between linear expansivity (α\alpha) and volume (cubical) expansivity (γ\gamma).
γ=3α\gamma = 3\alpha
For an isotropic solid, volume expansion occurs equally in three dimensions, making the volume coefficient three times the linear coefficient.
2
Substitute the given value of linear expansivity into the formula and calculate.
γ=3×(1.5×10−5 K−1)=4.5×10−5 K−1\gamma = 3 \times (1.5 \times 10^{-5}\text{ K}^{-1}) = 4.5 \times 10^{-5}\text{ K}^{-1}
Obtain the numeric value of volume expansivity.

Key Concept

Relationship between linear and volume expansivity of solids
Question 4Question →

A rectangular metal sheet has an initial area of 2.0 m22.0\text{ m}^2 at 20∘C20^\circ\text{C}. If the linear expansivity of the metal is 2.5×10−5 K−12.5 \times 10^{-5}\text{ K}^{-1}, what is the final temperature required for its area to increase by 0.005 m20.005\text{ m}^2?

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Answer: 70∘C70^\circ\text{C}

Answer

The final temperature required is 70∘C70^\circ\text{C}.
Area expansion is governed by ΔA=A0βΔT\Delta A = A_0 \beta \Delta T, where the area expansivity β=2α\beta = 2\alpha. Substituting A0=2.0 m2A_0 = 2.0\text{ m}^2, ΔA=0.005 m2\Delta A = 0.005\text{ m}^2, and β=5.0×10−5 K−1\beta = 5.0 \times 10^{-5}\text{ K}^{-1} gives a temperature rise ΔT=50∘C\Delta T = 50^\circ\text{C}. Adding the initial temperature of 20∘C20^\circ\text{C} yields a final temperature of 70∘C70^\circ\text{C}.

Step-by-Step Solution

1
Determine the area expansivity (superficial expansivity) β\beta from the linear expansivity α\alpha.
β=2α=2×2.5×10−5 K−1=5.0×10−5 K−1\beta = 2\alpha = 2 \times 2.5 \times 10^{-5}\text{ K}^{-1} = 5.0 \times 10^{-5}\text{ K}^{-1}
Area expansion depends on area expansivity, which is twice the linear expansivity.
2
Calculate the temperature change ΔT\Delta T using the formula ΔA=A0βΔT\Delta A = A_0 \beta \Delta T.
\(\Delta T = \frac{\Delta A}{A_0 \beta} = \frac{0.005\text{ m}^2}{2.0\text{ m}^2 \times 5.0 \times 10^{-5}\text{ K}^{-1}} = \frac{5 \times 10^{-3}}{1.0 \times 10^{-4}} = 50\text{ K}\)
Rearranging the expansion formula isolates the temperature change variable.
3
Calculate the final temperature T2T_2 by adding ΔT\Delta T to the initial temperature T1T_1.
T2=T1+ΔT=20∘C+50∘C=70∘CT_2 = T_1 + \Delta T = 20^\circ\text{C} + 50^\circ\text{C} = 70^\circ\text{C}
The final temperature is the sum of the initial temperature and the rise in temperature.

Key Concept

Thermal Expansion of Solids (Area Expansivity)
Estimated Time:1m 30s
Question 5Question →

A copper rod has an initial length of 100 cm100\text{ cm} at 25∘C25^\circ\text{C}. When its temperature is increased to 75∘C75^\circ\text{C}, its length becomes 100.085 cm100.085\text{ cm}. What is the area expansivity of copper in 10−5 K−110^{-5}\text{ K}^{-1}?

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Answer: 3.4

Answer

The area expansivity of copper is 3.4×10−5 K−13.4 \times 10^{-5}\text{ K}^{-1}, giving a numerical value of 3.43.4 in units of 10−5 K−110^{-5}\text{ K}^{-1}.
Linear expansivity is obtained as α=ΔLL0ΔT=0.085 cm100 cm×50 K=1.7×10−5 K−1\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085\text{ cm}}{100\text{ cm} \times 50\text{ K}} = 1.7 \times 10^{-5}\text{ K}^{-1}. Since area expansivity β\beta is related to linear expansivity by β=2α\beta = 2\alpha, multiplying by 22 yields 3.4×10−5 K−13.4 \times 10^{-5}\text{ K}^{-1}, or 3.43.4 in units of 10−5 K−110^{-5}\text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the temperature change
ΔT=75∘C−25∘C=50 K\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50\text{ K}
Thermal expansion depends directly on the change in temperature.
2
Find the change in length
ΔL=100.085 cm−100 cm=0.085 cm\Delta L = 100.085\text{ cm} - 100\text{ cm} = 0.085\text{ cm}
The expansion is the difference between the final length and original length.
3
Determine linear expansivity (α\alpha)
α=ΔLL0ΔT=0.085100×50=1.7×10−5 K−1\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085}{100 \times 50} = 1.7 \times 10^{-5}\text{ K}^{-1}
Linear expansivity defines fractional change in length per degree temperature change.
4
Compute area expansivity (β\beta)
β=2α=2×(1.7×10−5)=3.4×10−5 K−1\beta = 2\alpha = 2 \times (1.7 \times 10^{-5}) = 3.4 \times 10^{-5}\text{ K}^{-1}
Area (superficial) expansivity is equal to twice the linear expansivity.

Key Concept

Relationship between linear expansivity (α\alpha) and area expansivity (β=2α\beta = 2\alpha).
Question 6Question →

An aluminium rod of initial length 2.0 m2.0\text{ m} at 20∘C20^\circ\text{C} expands by 0.96 mm0.96\text{ mm} when heated. If the linear expansivity of aluminium is 2.4×10−5 K−12.4 \times 10^{-5}\text{ K}^{-1}, what is the rise in temperature of the rod?

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Answer: 20

Answer

The rise in temperature of the aluminium rod is 20 K20\text{ K}.
The fractional change in length depends on linear expansivity and temperature change through ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. Substituting the converted expansion ΔL=9.6×10−4 m\Delta L = 9.6 \times 10^{-4}\text{ m}, initial length L0=2.0 mL_0 = 2.0\text{ m}, and linear expansivity α=2.4×10−5 K−1\alpha = 2.4 \times 10^{-5}\text{ K}^{-1} gives ΔT=9.6×10−42.0×2.4×10−5=20 K\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times 2.4 \times 10^{-5}} = 20\text{ K}.

Step-by-Step Solution

1
Convert change in length from millimeters to meters
\Delta L = 9.6 \times 10^{-4}\text{ m}
Units must be consistent with initial length in meters.
2
Rearrange the linear thermal expansion formula \Delta L = L_0 \alpha \Delta T for temperature change \Delta T
\Delta T = \frac{\Delta L}{L_0 \alpha}
To isolate the unknown quantity \Delta T.
3
Substitute values into the rearranged formula and compute \Delta T
\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times (2.4 \times 10^{-5})} = 20\text{ K}
Evaluating the mathematical expression yields the required temperature rise.

Key Concept

Linear Expansivity and Thermal Expansion of Solids
Question 7Question →

A railway line is laid using steel rails, each of length 15 m15\text{ m}, at a temperature of 20∘C20^\circ\text{C}. What minimum gap, in millimetres (mm\text{mm}), must be left between consecutive rails so that they just touch without buckling when heated to 60∘C60^\circ\text{C}? [Linear expansivity of steel = 1.2×10−5 K−11.2 \times 10^{-5}\text{ K}^{-1}]

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Answer: 7.2

Answer

The minimum gap required between consecutive rails is 7.2 mm7.2\text{ mm}.
The expansion in length ΔL\Delta L is given by ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. Substituting the initial length L0=15 mL_0 = 15\text{ m}, linear expansivity α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}, and temperature change ΔT=40 K\Delta T = 40\text{ K} gives ΔL=7.2×10−3 m\Delta L = 7.2 \times 10^{-3}\text{ m}, which corresponds to 7.2 mm7.2\text{ mm}.

Step-by-Step Solution

1
Determine the change in temperature
\Delta T = 60^\circ\text{C} - 20^\circ\text{C} = 40\text{ K}
Thermal expansion is driven by the temperature difference between the final and initial states.
2
Set up the linear expansion formula
\Delta L = L_0 \alpha \Delta T
The linear expansion of a solid bar depends on its initial length, the material's linear expansivity, and the temperature change.
3
Calculate the expansion in metres and convert to millimetres
\Delta L = 15 \times (1.2 \times 10^{-5}) \times 40 = 7.2 \times 10^{-3}\text{ m} = 7.2\text{ mm}
Multiplying the value in metres by 10001000 yields the required measurement in millimetres.

Key Concept

Linear Expansivity and Thermal Expansion of Solids
Estimated Time:1m 30s
Question 8Question →

A solid metal block of mass 4.0 kg4.0\text{ kg} has a density of 8000 kg m−38000\text{ kg m}^{-3} at 20∘C20^\circ\text{C}. If the linear expansivity of the metal is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}, calculate the increase in volume of the block, in cm3\text{cm}^3, when its temperature is raised to 120∘C120^\circ\text{C}.

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Answer: 3

Answer

The increase in volume of the block is 3.0 cm33.0\text{ cm}^3.
First, the initial volume of the metal block is found using V0=mρ=4.0 kg8000 kg m−3=5.0×10−4 m3=500 cm3V_0 = \frac{m}{\rho} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3. Next, because a solid expands in three dimensions, the cubic expansivity is γ=3α=3×(2.0×10−5 K−1)=6.0×10−5 K−1\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}. The temperature change is ΔT=120∘C−20∘C=100 K\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}. Finally, the increase in volume is computed as ΔV=V0γΔT=500 cm3×(6.0×10−5 K−1)×100 K=3.0 cm3\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3.

Step-by-Step Solution

1
Calculate the initial volume of the metal block from its mass and density.
V0=mρ0=4.0 kg8000 kg m−3=5.0×10−4 m3=500 cm3V_0 = \frac{m}{\rho_0} = \frac{4.0\text{ kg}}{8000\text{ kg m}^{-3}} = 5.0 \times 10^{-4}\text{ m}^3 = 500\text{ cm}^3
Volume is equal to mass divided by density.
2
Convert linear expansivity (α\alpha) to volume expansivity (γ\gamma).
γ=3α=3×(2.0×10−5 K−1)=6.0×10−5 K−1\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}
For an isotropic solid, cubic expansivity is three times its linear expansivity.
3
Determine the change in temperature.
ΔT=120∘C−20∘C=100 K\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}
Thermal expansion is driven by the change in temperature.
4
Calculate the total volume expansion.
ΔV=V0γΔT=500 cm3×(6.0×10−5 K−1)×100 K=3.0 cm3\Delta V = V_0 \gamma \Delta T = 500\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 3.0\text{ cm}^3
Applying the thermal volume expansion formula ΔV=V0γΔT\Delta V = V_0 \gamma \Delta T.

Key Concept

Thermal Volume Expansion of Solids
Estimated Time:2m 0s
Question 9Question →

A metallic rod of initial length 2.0 m2.0\text{ m} experiences a temperature increase of 50 K50\text{ K}. If the linear expansivity of the metal is 1.5×10−5 K−11.5 \times 10^{-5}\text{ K}^{-1}, what is the expansion in length of the rod in millimetres?

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Answer: 1.5

Answer

The expansion in length of the rod is 1.5 mm.
The expansion in length is calculated using ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. Substituting the values L0=2.0 mL_0 = 2.0\text{ m}, α=1.5×10−5 K−1\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}, and ΔT=50 K\Delta T = 50\text{ K} gives ΔL=1.5×10−3 m\Delta L = 1.5 \times 10^{-3}\text{ m}, which equals 1.5 mm1.5\text{ mm}.

Step-by-Step Solution

1
Identify known variables from the problem statement.
L0=2.0 mL_0 = 2.0\text{ m}, ΔT=50 K\Delta T = 50\text{ K}, and α=1.5×10−5 K−1\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}.
Listing given physical quantities clarifies which thermal expansion formula to apply.
2
Calculate the change in length in metres using ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T.
ΔL=2.0 m×(1.5×10−5 K−1)×50 K=0.0015 m\Delta L = 2.0\text{ m} \times (1.5 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 0.0015\text{ m}.
Thermal expansion in one dimension is directly proportional to initial length, linear expansivity, and temperature change.
3
Convert the calculated expansion from metres to millimetres.
0.0015 m×1000 mm/m=1.5 mm0.0015\text{ m} \times 1000\text{ mm/m} = 1.5\text{ mm}.
The question explicitly requests the value in millimetres.

Key Concept

Linear thermal expansivity defines the fractional change in length per degree temperature change.
Question 10Question →

A steel rod and a brass rod are arranged such that the difference between their lengths remains constant at 10 cm10\text{ cm} at all temperatures. If the linear expansivity of steel is 1.2×10−5 K−11.2 \times 10^{-5}\text{ K}^{-1} and that of brass is 1.8×10−5 K−11.8 \times 10^{-5}\text{ K}^{-1}, what is the initial length of the steel rod in centimetres?

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Answer: 30

Answer

The initial length of the steel rod is 30 cm30\text{ cm}.
For the length difference between two rods to remain constant regardless of temperature change, both rods must undergo equal absolute expansion (\(\Delta L_1 = \Delta L_2\)). Since \(\Delta L = L_0 \alpha \Delta T\), this requires \(L_1 \alpha_1 = L_2 \alpha_2\). Substituting \(L_{\text{brass}} = L_{\text{steel}} - 10\text{ cm}\) and the given expansivity values gives \(1.2 \times 10^{-5} L_{\text{steel}} = 1.8 \times 10^{-5} (L_{\text{steel}} - 10)\), which simplifies to \(0.6 L_{\text{steel}} = 18\), giving \(L_{\text{steel}} = 30\text{ cm}\).

Step-by-Step Solution

1
Relate the condition for a constant difference in length to individual expansions
\(\Delta L_{\text{steel}} = \Delta L_{\text{brass}}\)
If the difference between the two lengths is constant across temperature changes, both rods must increase in length by the exact same amount for any given temperature change.
2
Apply the linear thermal expansion formula to both rods
\(L_{\text{steel}} \alpha_{\text{steel}} = L_{\text{brass}} \alpha_{\text{brass}}\)
Since \(\Delta L = L_0 \alpha \Delta T\), setting \(\Delta L_{\text{steel}} = \Delta L_{\text{brass}}\) gives \(L_{\text{steel}} \alpha_{\text{steel}} \Delta T = L_{\text{brass}} \alpha_{\text{brass}} \Delta T\). Cancelling \(\Delta T\) yields \(L_{\text{steel}} \alpha_{\text{steel}} = L_{\text{brass}} \alpha_{\text{brass}}\).
3
Substitute the length relationship into the equation
\(L_{\text{steel}} (1.2 \times 10^{-5}) = (L_{\text{steel}} - 10) (1.8 \times 10^{-5})\)
Because brass has a larger linear expansivity than steel, the brass rod must be shorter than the steel rod so that their products of length and expansivity remain equal, hence \(L_{\text{brass}} = L_{\text{steel}} - 10\text{ cm}\).
4
Solve for the length of the steel rod
\(L_{\text{steel}} = 30\text{ cm}\)
Dividing both sides by \(10^{-5}\) gives \(1.2 L_{\text{steel}} = 1.8 L_{\text{steel}} - 18\). Rearranging gives \(0.6 L_{\text{steel}} = 18\), which yields \(L_{\text{steel}} = \frac{18}{0.6} = 30\text{ cm}\).

Key Concept

Equal absolute linear expansion for constant length difference
Estimated Time:2m 0s
Question 11Question →

A thin flat metal plate has an initial surface area of 0.5 m20.5\text{ m}^2 at 10∘C10^\circ\text{C}. When the plate is heated to a final temperature of 110∘C110^\circ\text{C}, its surface area increases by 1.8×10−3 m21.8 \times 10^{-3}\text{ m}^2. What is the coefficient of cubical expansivity of the metal?

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Answer: 5.4×10−5 K−15.4 \times 10^{-5}\text{ K}^{-1}

Answer

5.4×10−5 K−15.4 \times 10^{-5}\text{ K}^{-1}
The correct answer is derived by first finding the area expansivity β=ΔAA0ΔT=3.6×10−5 K−1\beta = \frac{\Delta A}{A_0 \Delta T} = 3.6 \times 10^{-5}\text{ K}^{-1}. Since β=2α\beta = 2\alpha, the linear expansivity α=1.8×10−5 K−1\alpha = 1.8 \times 10^{-5}\text{ K}^{-1}. The cubical expansivity is γ=3α=5.4×10−5 K−1\gamma = 3\alpha = 5.4 \times 10^{-5}\text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the temperature change ΔT\Delta T
ΔT=110∘C−10∘C=100 K\Delta T = 110^\circ\text{C} - 10^\circ\text{C} = 100\text{ K}
Expansion calculations depend on the temperature increase.
2
Determine the coefficient of area expansivity β\beta
β=ΔAA0ΔT=1.8×10−30.5×100=3.6×10−5 K−1\beta = \frac{\Delta A}{A_0 \Delta T} = \frac{1.8 \times 10^{-3}}{0.5 \times 100} = 3.6 \times 10^{-5}\text{ K}^{-1}
The area expansion formula is ΔA=A0βΔT\Delta A = A_0 \beta \Delta T.
3
Calculate the coefficient of linear expansivity α\alpha
α=β2=3.6×10−52=1.8×10−5 K−1\alpha = \frac{\beta}{2} = \frac{3.6 \times 10^{-5}}{2} = 1.8 \times 10^{-5}\text{ K}^{-1}
Area expansivity is twice the linear expansivity (β=2α\beta = 2\alpha).
4
Calculate the coefficient of cubical expansivity γ\gamma
γ=3α=3×(1.8×10−5)=5.4×10−5 K−1\gamma = 3\alpha = 3 \times (1.8 \times 10^{-5}) = 5.4 \times 10^{-5}\text{ K}^{-1}
Cubical expansivity is three times the linear expansivity (γ=3α\gamma = 3\alpha).

Key Concept

Relationship between linear (α\alpha), area (β\beta), and volume (γ\gamma) expansivities: β=2α\beta = 2\alpha and γ=3α\gamma = 3\alpha.
Estimated Time:1m 30s
Question 12Question →

A cylindrical metal rivet has a diameter of 2.50 cm2.50\text{ cm} at a room temperature of 25∘C25^\circ\text{C}. It needs to be inserted into a hole of diameter 2.49 cm2.49\text{ cm} in a structural frame. By how many kelvins must the rivet be cooled so that its diameter shrinks to just match the diameter of the hole? (Linear expansivity of the metal is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}).

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Answer: 200

Answer

The rivet must be cooled by 200 K.
Thermal expansion or contraction of a linear dimension (such as diameter) is governed by Δd=d0αΔT\Delta d = d_0 \alpha \Delta T. Substituting Δd=−0.01 cm\Delta d = -0.01\text{ cm}, d0=2.50 cmd_0 = 2.50\text{ cm}, and α=2.0×10−5 K−1\alpha = 2.0 \times 10^{-5}\text{ K}^{-1} gives −0.01=2.50×(2.0×10−5)×ΔT-0.01 = 2.50 \times (2.0 \times 10^{-5}) \times \Delta T, leading to ΔT=−200 K\Delta T = -200\text{ K}. Hence, cooling by 200 K is required.

Step-by-Step Solution

1
Calculate the required change in diameter
\Delta d = 2.49\text{ cm} - 2.50\text{ cm} = -0.01\text{ cm}
The diameter of the rivet must decrease from 2.50 cm to 2.49 cm to fit into the hole.
2
Set up the linear expansion equation
\Delta d = d_0 \alpha \Delta T
Linear contraction/expansion applies directly to any linear dimension of a solid, including diameter.
3
Substitute given values into the equation
-0.01\text{ cm} = (2.50\text{ cm}) \times (2.0 \times 10^{-5}\text{ K}^{-1}) \times \Delta T
Substitute initial diameter, linear expansivity, and change in diameter.
4
Solve for the temperature change
\Delta T = \frac{-0.01}{5.0 \times 10^{-5}} = -200\text{ K}
Dividing the change in length by the product of initial length and linear expansivity yields the temperature change.

Key Concept

Thermal Contraction and Linear Expansivity of Solids
Question 13Question →

A cylindrical brass sleeve has an internal diameter of 5.000 cm5.000\text{ cm} at a room temperature of 20∘C20^\circ\text{C}. It is to be shrink-fitted onto a solid shaft of diameter 5.012 cm5.012\text{ cm} (also at 20∘C20^\circ\text{C}). Assuming the linear expansivity of brass is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}, what is the minimum temperature, in ∘C^\circ\text{C}, to which the brass sleeve must be heated so that it just slips over the shaft?

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Answer: 140

Answer

140 °C
The required expansion in internal diameter is Δd=5.012 cm−5.000 cm=0.012 cm\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}. Using the linear expansion relation Δd=d0αΔT\Delta d = d_0 \alpha \Delta T, the required temperature change is ΔT=0.0125.000×2.0×10−5=120∘C\Delta T = \frac{0.012}{5.000 \times 2.0 \times 10^{-5}} = 120^\circ\text{C}. Adding this to the initial temperature of 20∘C20^\circ\text{C} gives a final minimum temperature of 140∘C140^\circ\text{C}.

Step-by-Step Solution

1
Determine the required increase in internal diameter (Δd\Delta d) of the brass sleeve
Δd=5.012 cm−5.000 cm=0.012 cm\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}
The sleeve's internal diameter must expand until it equals the shaft diameter.
2
Apply the linear expansion formula Δd=d0αΔT\Delta d = d_0 \alpha \Delta T to find the temperature rise ΔT\Delta T
ΔT=0.012 cm5.000 cm×2.0×10−5 K−1=0.0121.0×10−4=120 K\Delta T = \frac{0.012\text{ cm}}{5.000\text{ cm} \times 2.0 \times 10^{-5}\text{ K}^{-1}} = \frac{0.012}{1.0 \times 10^{-4}} = 120\text{ K}
Linear dimensions such as diameter expand in direct proportion to the linear expansivity coefficient α\alpha.
3
Calculate the final temperature T2T_2
T2=T1+ΔT=20∘C+120∘C=140∘CT_2 = T_1 + \Delta T = 20^\circ\text{C} + 120^\circ\text{C} = 140^\circ\text{C}
The final temperature is found by adding the temperature increase to the initial temperature.

Key Concept

Linear Expansivity and One-Dimensional Expansion of Curved Boundaries
Question 14Question →

A solid metal sphere has an initial volume of 1000 cm31000\text{ cm}^3 at 20∘C20^\circ\text{C}. If it is heated to a final temperature of 70∘C70^\circ\text{C} and the linear expansivity of the metal is 2.0×10−5 K−12.0 \times 10^{-5}\text{ K}^{-1}, what is the increase in the volume of the sphere?

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Answer: 3.0 cm33.0\text{ cm}^3

Answer

The increase in the volume of the sphere is 3.0 cm33.0\text{ cm}^3.
The value of 3.0 cm33.0\text{ cm}^3 is correct because the volume expansion requires the cubical expansivity γ=3α=6.0×10−5 K−1\gamma = 3\alpha = 6.0 \times 10^{-5}\text{ K}^{-1}. Multiplying this coefficient by the initial volume (1000 cm31000\text{ cm}^3) and the temperature change (50 K50\text{ K}) yields ΔV=1000×6.0×10−5×50=3.0 cm3\Delta V = 1000 \times 6.0 \times 10^{-5} \times 50 = 3.0\text{ cm}^3.

Step-by-Step Solution

1
Calculate the temperature change (ΔT\Delta T).
ΔT=70∘C−20∘C=50∘C=50 K\Delta T = 70^\circ\text{C} - 20^\circ\text{C} = 50^\circ\text{C} = 50\text{ K}
Thermal expansion depends on the change in temperature rather than the initial or final temperature alone.
2
Determine the cubical (volume) expansivity (γ\gamma) from linear expansivity (α\alpha).
γ=3α=3×(2.0×10−5 K−1)=6.0×10−5 K−1\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}
For an isotropic solid, volume expands in three orthogonal dimensions, making cubical expansivity equal to three times linear expansivity.
3
Calculate the increase in volume (ΔV\Delta V).
ΔV=V1γΔT=1000 cm3×(6.0×10−5 K−1)×50 K=3.0 cm3\Delta V = V_1 \gamma \Delta T = 1000\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 3.0\text{ cm}^3
Substitute initial volume, volume expansivity, and temperature change into the volume expansion formula.

Key Concept

Relationship between linear and volume expansivity (γ=3α\gamma = 3\alpha) and application of the volume expansion formula.
Estimated Time:1m 30s
Question 15Question →

A rectangular metallic sheet with a linear expansivity of 1.8×10−5 K−11.8 \times 10^{-5}\text{ K}^{-1} experiences a temperature rise of 50 K50\text{ K}. If the increase in its surface area is 0.90 cm20.90\text{ cm}^2, what was the initial surface area of the sheet in cm2\text{cm}^2?

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Answer: 500

Answer

The initial surface area of the metallic sheet is 500 cm2500\text{ cm}^2.
The initial area is found by converting linear expansivity to area expansivity (\beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}) and substituting into the area expansion relation \Delta A = A_0 \beta \Delta T, giving A_0 = \frac{0.90}{3.6 \times 10^{-5} \times 50} = 500\text{ cm}^2$.

Step-by-Step Solution

1
Calculate the area (superficial) expansivity (\beta)
\beta = 2\alpha = 2 \times 1.8 \times 10^{-5}\text{ K}^{-1} = 3.6 \times 10^{-5}\text{ K}^{-1}
Surface area expansion depends on area expansivity, which is twice the linear expansivity for an isotropic solid.
2
Formulate the thermal area expansion equation
\Delta A = A_0 \beta \Delta T
The fractional change in area is directly proportional to the area expansivity and the temperature change.
3
Rearrange the formula to solve for the initial surface area (A_0)
A_0 = \frac{\Delta A}{\beta \Delta T}
Isolating the required unknown quantity.
4
Substitute the known numerical values and compute
A_0 = \frac{0.90\text{ cm}^2}{(3.6 \times 10^{-5}\text{ K}^{-1})(50\text{ K})} = \frac{0.90}{1.8 \times 10^{-3}} = 500\text{ cm}^2
Evaluating the expression yields the exact initial surface area.

Key Concept

Relationship between Linear Expansivity and Area Expansivity

Alternative Method

Calculate fractional area expansion per kelvin: \beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}.Totalfractionalexpansionfor. Total fractional expansion for 50\text{ K}is is 3.6 \times 10^{-5} \times 50 = 0.0018 .Theninitialarea. Then initial area A_0 = \frac{0.90}{0.0018} = 500\text{ cm}^2$.
Estimated Time:1m 30s
Question 16Question →

An aluminum electric cable suspended between two transmission poles has a length of 100 m100\text{ m} at an initial morning temperature of 20∘C20^\circ\text{C}. By afternoon, the cable temperature increases to 50∘C50^\circ\text{C}. Given that the linear expansivity of aluminum is 2.3×10−5 K−12.3 \times 10^{-5}\text{ K}^{-1}, what is the increase in the length of the cable in centimeters?

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Answer: 6.9

Answer

The increase in the length of the cable is 6.9 cm6.9\text{ cm}.
Applying the formula for linear expansion ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T, where L0=100 mL_0 = 100\text{ m}, α=2.3×10−5 K−1\alpha = 2.3 \times 10^{-5}\text{ K}^{-1}, and ΔT=30 K\Delta T = 30\text{ K}, gives ΔL=0.069 m\Delta L = 0.069\text{ m}. Converting this to centimeters yields 6.9 cm6.9\text{ cm}.

Step-by-Step Solution

1
Determine the temperature change
\Delta T = 30\text{ K}
Temperature change is the difference between final and initial temperatures: 50∘C−20∘C=30 K50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}.
2
Calculate expansion in meters using the linear expansivity formula
\Delta L = 0.069\text{ m}
\Delta L = L_0 \alpha \Delta T = 100 \times (2.3 \times 10^{-5}) \times 30 = 0.069\text{ m}.
3
Convert the change in length to centimeters
\Delta L = 6.9\text{ cm}
Since 1 m=100 cm1\text{ m} = 100\text{ cm}, multiply 0.069 m0.069\text{ m} by 100100 to obtain 6.9 cm6.9\text{ cm}.

Key Concept

Linear expansivity of solid conductors
Estimated Time:1m 30s
Thermal Expansion of Solids (Linear, Area, and Volume Expansivity) Practice Questions — JAMB UTME | Examkin