Thermal Physics Practice Questions — JAMB UTME — Page 4

Question 61Question →

A resistance thermometer registers a resistance of $5.0\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $25.0\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When placed in a heated liquid bath, the resistance measured is $30.0\,\Omega$ . What is the temperature of the bath on the Kelvin scale?

125\,\text{K}

398\,\text{K}

423\,\text{K}

373\,\text{K}

Show answer & explanation

Answer: $398\,\text{K}$

Answer

The temperature of the bath on the Kelvin scale is

398\,\text{K}

.

Using the linear interpolation formula for a thermometric property

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

, substituting

R_0 = 5.0\,\Omega

,

R_{100} = 25.0\,\Omega

, and

R_\theta = 30.0\,\Omega

yields

\theta = \frac{25.0}{20.0} \times 100 = 125^\circ\text{C}

. Converting to absolute thermodynamic temperature gives

T = 125 + 273 = 398\,\text{K}

.

Step-by-Step Solution

1

Calculate the temperature on the Celsius scale using linear interpolation of thermometric property.

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{30.0 - 5.0}{25.0 - 5.0} \times 100^\circ\text{C} = \frac{25.0}{20.0} \times 100^\circ\text{C} = 125^\circ\text{C}

The change in resistance is directly proportional to the temperature change between fixed points.

2

Convert the temperature from degrees Celsius to Kelvins.

T = \theta + 273 = 125 + 273 = 398\,\text{K}

Absolute temperature in Kelvin is obtained by adding 273 to the temperature in degrees Celsius.

Key Concept

Temperature Scale Interpolation and Kelvin Conversion

Estimated Time:1m 30s

Question 62Question →

The length of the mercury column in an uncalibrated thermometer is $4.0\,\text{cm}$ at the ice point ( $0^\circ\text{C}$ ) and $24.0\,\text{cm}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature in degrees Celsius when the length of the mercury column is $19.0\,\text{cm}$ ?

Show answer & explanation

Answer: 75

Answer

The temperature corresponding to a mercury column length of

19.0\,\text{cm}

is

75^\circ\text{C}

.

The temperature on the Celsius scale is determined by the ratio of the length change above the ice point to the total length change between the ice and steam points:

T = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

. Substituting

L_0 = 4.0\,\text{cm}

,

L_{100} = 24.0\,\text{cm}

, and

L_T = 19.0\,\text{cm}

gives

T = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}

.

Step-by-Step Solution

1

Identify given thermometric length values at fixed points and at the unknown temperature

L_0 = 4.0\,\text{cm}

,

L_{100} = 24.0\,\text{cm}

, and

L_T = 19.0\,\text{cm}

These represent the length at the lower fixed point (

0^\circ\text{C}

), upper fixed point (

100^\circ\text{C}

), and intermediate temperature

T

respectively.

2

Set up the linear interpolation equation on the Celsius scale

T = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

Thermometric expansion is assumed to vary linearly with temperature over the operational range.

3

Substitute the given values and perform arithmetic calculation

T = \frac{19.0 - 4.0}{24.0 - 4.0} \times 100 = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}

Simplifying

\frac{15.0}{20.0}

gives

0.75

, which multiplied by

100

equals

75

.

Key Concept

Temperature measurement using linear variation of thermometric properties

Question 63Question →

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

Show answer & explanation

Answer: 20

Answer

The rise in temperature of the aluminium rod is

20\text{ K}

.

The fractional change in length depends on linear expansivity and temperature change through

\Delta L = L_0 \alpha \Delta T

. Substituting the converted expansion

\Delta L = 9.6 \times 10^{-4}\text{ m}

, initial length

L_0 = 2.0\text{ m}

, and linear expansivity

\alpha = 2.4 \times 10^{-5}\text{ K}^{-1}

gives

\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 64Question →

A glass flask of volume $1000\text{ cm}^3$ is filled completely with mercury at a temperature of $10^\circ\text{C}$ . The linear expansivity of the glass is $9.0 \times 10^{-6}\text{ K}^{-1}$ and the real cubic expansivity of mercury is $1.8 \times 10^{-4}\text{ K}^{-1}$ . What volume of mercury (in $\text{cm}^3$ ) will overflow when the system is heated to $110^\circ\text{C}$ ?

Show answer & explanation

Answer: 15.3

Answer

The volume of mercury that overflows is

15.3\text{ cm}^3

.

The apparent expansion of the liquid equals its real expansion minus the expansion of the container. Since

\gamma_v = 3\alpha = 2.7 \times 10^{-5}\text{ K}^{-1} = 0.27 \times 10^{-4}\text{ K}^{-1}

, the apparent cubic expansivity is

\gamma_a = 1.8 \times 10^{-4} - 0.27 \times 10^{-4} = 1.53 \times 10^{-4}\text{ K}^{-1}

. Multiplying by initial volume (

1000\text{ cm}^3

) and temperature change (

100\text{ K}

) yields an overflow volume of

15.3\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the volume expansivity of the glass vessel (

\gamma_v

)

\gamma_v = 3 \times 9.0 \times 10^{-6}\text{ K}^{-1} = 2.7 \times 10^{-5}\text{ K}^{-1} = 0.27 \times 10^{-4}\text{ K}^{-1}

The volumetric (cubic) expansivity of a solid container is three times its linear expansivity.

2

Determine the apparent cubic expansivity of mercury (

\gamma_a

)

\gamma_a = \gamma_r - \gamma_v = 1.8 \times 10^{-4} - 0.27 \times 10^{-4} = 1.53 \times 10^{-4}\text{ K}^{-1}

The apparent expansion of a liquid accounts for both the expansion of the liquid itself and the expansion of the containing vessel.

3

Calculate the overflow volume (apparent expansion

\Delta V_a

)

\Delta V_a = V_0 \times \gamma_a \times \Delta T = 1000 \times 1.53 \times 10^{-4} \times 100 = 15.3\text{ cm}^3

The volume of liquid that overflows corresponds directly to its apparent volume increase.

Key Concept

Real and Apparent Cubical Expansivity of Liquids

Estimated Time:1m 30s

Question 65Question →

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

Show answer & explanation

Answer: 7.2

Answer

The minimum gap required between consecutive rails is

7.2\text{ mm}

.

The expansion in length

\Delta L

is given by

\Delta L = L_0 \alpha \Delta T

. Substituting the initial length

L_0 = 15\text{ m}

, linear expansivity

\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}

, and temperature change

\Delta T = 40\text{ K}

gives

\Delta L = 7.2 \times 10^{-3}\text{ m}

, which corresponds to

7.2\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

1000

yields the required measurement in millimetres.

Key Concept

Linear Expansivity and Thermal Expansion of Solids

Estimated Time:1m 30s

Question 66Question →

A brass container with an initial capacity of $500\text{ cm}^3$ at $20^\circ\text{C}$ is filled completely with ethanol. When the container and its contents are uniformly heated to $70^\circ\text{C}$ , a volume of $12\text{ cm}^3$ of ethanol overflows from the container. Given that the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , calculate the real cubic expansivity of ethanol in units of $10^{-4}\text{ K}^{-1}$ .

Show answer & explanation

Answer: 5.4

Answer

The real cubic expansivity of ethanol is

5.4 \times 10^{-4}\text{ K}^{-1}

, which corresponds to a value of

5.4

in units of

10^{-4}\text{ K}^{-1}

.

The real volume increase of a liquid consists of the apparent expansion (observed overflow volume) plus the volume expansion of the container itself. By finding the apparent cubic expansivity

\gamma_a = \frac{12}{500 \times 50} = 4.8 \times 10^{-4}\text{ K}^{-1}

and adding the vessel's cubic expansivity

\gamma_v = 3 \times 2.0 \times 10^{-5} = 0.6 \times 10^{-4}\text{ K}^{-1}

, we obtain the real cubic expansivity

\gamma_r = 5.4 \times 10^{-4}\text{ K}^{-1}

, which yields

5.4

in the required units.

Step-by-Step Solution

1

Determine the temperature increase of the system

\Delta T = 70^\circ\text{C} - 20^\circ\text{C} = 50\text{ K}

The thermal expansion is driven by the change in temperature.

2

Calculate the apparent cubic expansivity of ethanol (\gamma_a)

\gamma_a = \frac{\Delta V_{overflow}}{V_0 \cdot \Delta T} = \frac{12\text{ cm}^3}{500\text{ cm}^3 \times 50\text{ K}} = 4.8 \times 10^{-4}\text{ K}^{-1}

The overflow volume represents the apparent volume increase of the liquid relative to the expanding vessel.

3

Calculate the cubic expansivity of the brass vessel (\gamma_v)

\gamma_v = 3 \times \alpha_{brass} = 3 \times 2.0 \times 10^{-5}\text{ K}^{-1} = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}

Cubic expansivity of a solid vessel is three times its linear expansivity.

4

Compute the real cubic expansivity of ethanol (\gamma_r)

\gamma_r = \gamma_a + \gamma_v = 4.8 \times 10^{-4}\text{ K}^{-1} + 0.6 \times 10^{-4}\text{ K}^{-1} = 5.4 \times 10^{-4}\text{ K}^{-1}

The real expansion of a liquid is the sum of its apparent expansion and the expansion of the containing vessel.

Key Concept

Relationship between Real and Apparent Cubic Expansivity of Liquids

Question 67Question →

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

Show answer & explanation

Answer: 3

Answer

The increase in volume of the block is

3.0\text{ cm}^3

.

First, the initial volume of the metal block is found using

V_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

\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

\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}

. Finally, the increase in volume is computed as

\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.

V_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 (

\alpha

) to volume expansivity (

\gamma

).

\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.

\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.

\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

\Delta V = V_0 \gamma \Delta T

.

Key Concept

Thermal Volume Expansion of Solids

Estimated Time:2m 0s

Question 68Question →

A faulty liquid-in-glass thermometer registers a reading of $5^\circ\text{C}$ at the melting ice point and $95^\circ\text{C}$ at the steam point of pure water under standard atmospheric pressure. What is the actual temperature in degrees Celsius when this thermometer registers a reading of $41^\circ\text{C}$ ?

36^\circ\text{C}

40^\circ\text{C}

41^\circ\text{C}

45.6^\circ\text{C}

Show answer & explanation

Answer: $40^\circ\text{C}$

Answer

The actual temperature is

40^\circ\text{C}

.

The correct answer is obtained by setting up the linear interpolation formula for thermometric property values:

\theta = \frac{X_\theta - X_0}{X_{100} - X_0} \times 100

. Substituting

X_0 = 5

,

X_{100} = 95

, and

X_\theta = 41

gives

\theta = \frac{36}{90} \times 100 = 40^\circ\text{C}

.

Step-by-Step Solution

1

Determine the fundamental interval of the faulty thermometer.

Fundamental interval

= 95^\circ\text{C} - 5^\circ\text{C} = 90\,\text{divisions}

.

The total interval between the lower fixed point and upper fixed point represents

100^\circ\text{C}

on the standard Celsius scale.

2

Calculate the measured change from the ice point.

Measured difference

= 41^\circ\text{C} - 5^\circ\text{C} = 36\,\text{divisions}

.

The zero error of

+5^\circ\text{C}

must be subtracted from the observed reading.

3

Apply the linear scale interpolation formula to find the actual temperature

\theta

.

\theta = \frac{41 - 5}{95 - 5} \times 100^\circ\text{C} = \frac{36}{90} \times 100^\circ\text{C} = 40^\circ\text{C}

.

The ratio of the measured interval to the total fundamental interval equals the true fraction of

100^\circ\text{C}

.

Key Concept

Linear interpolation on non-standard or faulty thermometer scales using fixed points.

Question 69Question →

A metal container with an initial volume of $500\text{ cm}^3$ is completely filled with an organic liquid at $15^\circ\text{C}$ . The real cubic expansivity of the liquid is $4.0 \times 10^{-4}\text{ K}^{-1}$ . When the container and liquid are heated together to a final temperature of $65^\circ\text{C}$ , exactly $8.5\text{ cm}^3$ of the liquid overflows. What is the linear expansivity of the metal container, in units of $10^{-5}\text{ K}^{-1}$ ?

Show answer & explanation

Answer: 2

Answer

The linear expansivity of the metal container is

2.0 \times 10^{-5}\text{ K}^{-1}

, which corresponds to a numerical value of

2.0

in units of

10^{-5}\text{ K}^{-1}

.

The apparent expansivity of the liquid is derived from the overflow volume (8.5 cm^3 / (500 cm^3 * 50 K) = 3.4 x 10^-4 K^-1). Subtracting this from the liquid's real expansivity (4.0 x 10^-4 K^-1) yields the container's cubic expansivity of 0.6 x 10^-4 K^-1 (or 6.0 x 10^-5 K^-1). Dividing by 3 gives the linear expansivity of the container material as 2.0 x 10^-5 K^-1.

Step-by-Step Solution

1

Determine the temperature increase

\Delta T = 50\text{ K}

Temperature change drives the thermal expansion process.

2

Compute the apparent cubic expansivity of the liquid

\gamma_a = 3.4 \times 10^{-4}\text{ K}^{-1}

Apparent expansion is measured directly from the liquid overflow relative to initial volume and temperature rise.

3

Calculate the cubic expansivity of the metal container

\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1}

The difference between real cubic expansivity of the liquid and its apparent cubic expansivity equals the cubic expansivity of the container.

4

Calculate the linear expansivity of the container material

\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}

For isotropic solids, volume expansivity is three times linear expansivity (\gamma_v = 3\alpha).

Key Concept

Thermal expansion of liquids in expanding vessels: Real expansivity equals apparent expansivity plus vessel cubic expansivity (\gamma_r = \gamma_a + 3\alpha).

Estimated Time:2m 0s

Question 70Question →

A platinum resistance thermometer has a resistance of $10\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $50\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When placed in a warm liquid bath, an uncalibrated digital ohmmeter reads $38\,\Omega$ . If the ohmmeter has a known positive zero error of $+4\,\Omega$ (indicating $4\,\Omega$ above the true resistance), what is the actual temperature of the liquid bath on the absolute scale in kelvins?

333\,\text{K}

343\,\text{K}

353\,\text{K}

60\,\text{K}

Show answer & explanation

Answer: $333\,\text{K}$

Answer

The actual temperature of the liquid bath on the absolute scale is

333\,\text{K}

.

Subtracting the zero error of

+4\,\Omega

from the raw meter reading of

38\,\Omega

yields the true thermometric resistance of

34\,\Omega

. Applying the temperature formula

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

gives

\theta = \frac{34 - 10}{50 - 10} \times 100 = 60^\circ\text{C}

. Converting to absolute temperature gives

T = 60 + 273 = 333\,\text{K}

.

Step-by-Step Solution

1

Correct the measured resistance for zero error

True resistance

R_\theta = 38\,\Omega - 4\,\Omega = 34\,\Omega

A positive zero error means the meter reads higher than the true value, so the zero error must be subtracted.

2

Calculate the temperature on the Celsius scale using linear interpolation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{34 - 10}{50 - 10} \times 100 = \frac{24}{40} \times 100 = 60^\circ\text{C}

The thermometric property varies linearly between the fixed points.

3

Convert the Celsius temperature to the thermodynamic (absolute) Kelvin scale

T = \theta + 273 = 60 + 273 = 333\,\text{K}

The conversion from Celsius to Kelvin requires adding

273

(or

273.15

).

Key Concept

Linear interpolation of thermometric properties with instrument zero error correction

Estimated Time:2m 0s

Question 71Question →

A glass vessel with a linear expansivity of $9.0 \times 10^{-6} \text{ K}^{-1}$ has a volume capacity of $300 \text{ cm}^3$ at $15^\circ\text{C}$ . It is completely filled with a liquid at this temperature. When the system is uniformly heated to $65^\circ\text{C}$ , a volume of $4.5 \text{ cm}^3$ of the liquid overflows. What is the real cubic expansivity of the liquid?

3.27 \times 10^{-4} \text{ K}^{-1}

3.09 \times 10^{-4} \text{ K}^{-1}

3.00 \times 10^{-4} \text{ K}^{-1}

2.73 \times 10^{-4} \text{ K}^{-1}

Show answer & explanation

Answer: $3.27 \times 10^{-4} \text{ K}^{-1}$

Answer

The real cubic expansivity of the liquid is

3.27 \times 10^{-4} \text{ K}^{-1}

.

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubical expansivity of the containing vessel. The apparent cubic expansivity is

\frac{4.5}{300 \times 50} = 3.0 \times 10^{-4} \text{ K}^{-1}

. The cubical expansivity of the vessel is

3 \times 9.0 \times 10^{-6} = 2.7 \times 10^{-5} \text{ K}^{-1}

. Summing these yields

3.27 \times 10^{-4} \text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

) and the apparent cubic expansivity (

\gamma_a

) of the liquid.

\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K}

.

\gamma_a = \frac{V_{\text{overflow}}}{V_0 \cdot \Delta T} = \frac{4.5}{300 \times 50} = 3.0 \times 10^{-4} \text{ K}^{-1}

.

Apparent expansion measures the overflow relative to initial volume and temperature rise.

2

Calculate the cubical expansivity of the glass vessel (

\gamma_v

).

\gamma_v = 3 \alpha_v = 3 \times (9.0 \times 10^{-6} \text{ K}^{-1}) = 2.7 \times 10^{-5} \text{ K}^{-1} = 0.27 \times 10^{-4} \text{ K}^{-1}

.

The volume (cubical) expansivity of an isotropic solid container is three times its linear expansivity.

3

Determine the real cubic expansivity of the liquid (

\gamma_r

) using the relation

\gamma_r = \gamma_a + \gamma_v

.

\gamma_r = 3.0 \times 10^{-4} \text{ K}^{-1} + 0.27 \times 10^{-4} \text{ K}^{-1} = 3.27 \times 10^{-4} \text{ K}^{-1}

.

Real volume expansion equals the observed apparent expansion plus the expansion of the container.

Key Concept

Relationship between real expansivity, apparent expansivity, and vessel expansivity:

\gamma_r = \gamma_a + \gamma_v

, where

\gamma_v = 3\alpha

.

Question 72Question →

A liquid has a real cubic expansivity of $5.0 \times 10^{-4} \text{ K}^{-1}$ . When this liquid is heated inside a metallic vessel, its apparent cubic expansivity is determined to be $4.1 \times 10^{-4} \text{ K}^{-1}$ . What is the linear expansivity of the metallic vessel in $10^{-5} \text{ K}^{-1}$ ?

Show answer & explanation

Answer: 3

Answer

The linear expansivity of the metallic vessel is

3.0 \times 10^{-5} \text{ K}^{-1}

.

Real cubic expansivity of a liquid accounts for both the expansion of the liquid relative to the container and the expansion of the container itself:

\gamma_r = \gamma_a + \gamma_v

. Subtracting the apparent cubic expansivity (

4.1 \times 10^{-4} \text{ K}^{-1}

) from the real cubic expansivity (

5.0 \times 10^{-4} \text{ K}^{-1}

) gives the vessel's cubic expansivity of

0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}

. Dividing this value by 3 gives the vessel's linear expansivity:

\alpha = 3.0 \times 10^{-5} \text{ K}^{-1}

.

Step-by-Step Solution

1

Find the cubic expansivity of the vessel (

\gamma_v

)

\gamma_v = \gamma_r - \gamma_a = 5.0 \times 10^{-4} - 4.1 \times 10^{-4} = 0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the containing vessel (

\gamma_r = \gamma_a + \gamma_v

).

2

Calculate the linear expansivity of the vessel (

\alpha

)

\alpha = \frac{\gamma_v}{3} = \frac{9.0 \times 10^{-5}}{3} = 3.0 \times 10^{-5} \text{ K}^{-1}

For isotropic solids, volume (cubic) expansivity is three times the linear expansivity (

\gamma_v = 3\alpha

).

Key Concept

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel expansivity

Question 73Question →

A liquid has a real cubic expansivity of $7.5 \times 10^{-4}\text{ K}^{-1}$ . It is heated inside a metal vessel whose material has a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ . What is the ratio of the real cubic expansivity of the liquid to its apparent cubic expansivity?

\frac{25}{23}

\frac{75}{73}

\frac{25}{27}

\frac{75}{77}

Show answer & explanation

Answer: $\frac{25}{23}$

Answer

The ratio of the real cubic expansivity of the liquid to its apparent cubic expansivity is $\frac{25}{23}$.

The real cubic expansivity of a liquid $\gamma_r$ is related to its apparent cubic expansivity $\gamma_a$ and the cubic expansivity of the container $\gamma_v$ by the equation $\gamma_r = \gamma_a + \gamma_v$. For a vessel made of material with linear expansivity $\alpha$, $\gamma_v = 3\alpha$. Substituting $\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}$ yields $\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}$. The apparent expansivity is $\gamma_a = 7.5 \times 10^{-4} - 0.6 \times 10^{-4} = 6.9 \times 10^{-4}\text{ K}^{-1}$. The ratio $\frac{\gamma_r}{\gamma_a}$ is therefore $\frac{7.5 \times 10^{-4}}{6.9 \times 10^{-4}} = \frac{25}{23}$.

Step-by-Step Solution

1

Calculate the cubic expansivity of the metal vessel ($\gamma_v$) from its linear expansivity ($\alpha$).

$\gamma_v = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}$

The volume expansion coefficient of a solid vessel is three times its linear expansion coefficient.

2

Determine the apparent cubic expansivity ($\gamma_a$) using the relationship between real expansivity, apparent expansivity, and vessel expansivity.

$\gamma_a = \gamma_r - \gamma_v = 7.5 \times 10^{-4}\text{ K}^{-1} - 0.6 \times 10^{-4}\text{ K}^{-1} = 6.9 \times 10^{-4}\text{ K}^{-1}$

Real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the containing vessel.

3

Compute the ratio of real cubic expansivity to apparent cubic expansivity ($\frac{\gamma_r}{\gamma_a}$).

$\frac{\gamma_r}{\gamma_a} = \frac{7.5 \times 10^{-4}}{6.9 \times 10^{-4}} = \frac{75}{69} = \frac{25}{23}$

Dividing the given real cubic expansivity by the calculated apparent cubic expansivity gives the simplified fraction.

Key Concept

Thermal Expansion of Liquids: Relationship between Real and Apparent Expansivity

Estimated Time:1m 30s

Question 74Question →

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

Show answer & explanation

Answer: 1.5

Answer

The expansion in length of the rod is 1.5 mm.

The expansion in length is calculated using

\Delta L = L_0 \alpha \Delta T

. Substituting the values

L_0 = 2.0\text{ m}

,

\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}

, and

\Delta T = 50\text{ K}

gives

\Delta L = 1.5 \times 10^{-3}\text{ m}

, which equals

1.5\text{ mm}

.

Step-by-Step Solution

1

Identify known variables from the problem statement.

L_0 = 2.0\text{ m}

,

\Delta T = 50\text{ K}

, and

\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

\Delta L = L_0 \alpha \Delta T

.

\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\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 75Question →

A gas cylinder fitted with a pressure relief valve contains a fixed mass of an ideal gas at an initial absolute pressure of $2.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The valve is designed to open when the internal gauge pressure exceeds $3.20 \times 10^5\text{ Pa}$ . Assuming the atmospheric pressure is $1.00 \times 10^5\text{ Pa}$ , at what temperature will the relief valve open?

231^\circ\text{C}

45.4^\circ\text{C}

111^\circ\text{C}

34.6^\circ\text{C}

Show answer & explanation

Answer: $231^\circ\text{C}$

Answer

231^\circ\text{C}

The correct answer is

231^\circ\text{C}

. To solve this, first convert the initial temperature to Kelvin:

T_1 = 27 + 273 = 300\text{ K}

. Next, calculate the final absolute pressure by adding the atmospheric pressure to the given gauge pressure:

P_2 = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}

. Applying Gay-Lussac's Law (

\frac{P_1}{T_1} = \frac{P_2}{T_2}

) gives

T_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}

. Converting back to Celsius gives

504 - 273 = 231^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin and determine the initial absolute pressure.

T_1 = 27 + 273 = 300\text{ K}

,

P_1 = 2.50 \times 10^5\text{ Pa}

.

Gas laws require thermodynamic (absolute) temperature in Kelvin.

2

Calculate the final total absolute pressure (

P_2

) at which the valve opens.

P_2 = P_{\text{gauge}} + P_{\text{atm}} = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}

.

Gauge pressure measures the pressure difference relative to atmospheric pressure; gas laws use absolute pressure.

3

Apply Gay-Lussac's (Pressure) Law at constant volume to find the final Kelvin temperature (

T_2

).

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies T_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}

.

For a fixed volume of ideal gas, absolute pressure is directly proportional to absolute temperature.

4

Convert the final temperature back to degrees Celsius.

t_2 = 504 - 273 = 231^\circ\text{C}

.

The question asks for the temperature in degrees Celsius.

Key Concept

Pressure Law (Gay-Lussac's Law) and Absolute vs Gauge Pressure

Estimated Time:2m 0s

Question 76Question →

A liquid has a density of $840\text{ kg m}^{-3}$ at $10^\circ\text{C}$ . It is heated inside a container whose material has a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ . If the measured apparent cubic expansivity of the liquid in this container is $4.4 \times 10^{-4}\text{ K}^{-1}$ , what is the density of the liquid at $110^\circ\text{C}$ in $\text{kg m}^{-3}$ ?

Show answer & explanation

Answer: 800

Answer

800 kg m^-3

To find the density of the liquid at the elevated temperature, we must use its real cubic expansivity

\gamma_r

. The vessel expands with a cubic expansivity of

\gamma_v = 3\alpha = 6.0 \times 10^{-5}\text{ K}^{-1}

. The real cubic expansivity of the liquid is

\gamma_r = \gamma_a + \gamma_v = 4.4 \times 10^{-4} + 0.6 \times 10^{-4} = 5.0 \times 10^{-4}\text{ K}^{-1}

. Using the density variation formula

\rho_2 = \frac{\rho_1}{1 + \gamma_r \Delta T}

, we evaluate

\frac{840}{1 + (5.0 \times 10^{-4} \times 100)} = \frac{840}{1.05} = 800\text{ kg m}^{-3}

.

Step-by-Step Solution

1

Calculate the cubic expansivity of the container material

\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1}

The volume (cubic) expansivity of a solid container is three times its linear expansivity (

\gamma_v = 3\alpha

).

2

Determine the real cubic expansivity of the liquid

\gamma_r = 5.0 \times 10^{-4}\text{ K}^{-1}

The real cubic expansivity is the sum of the apparent cubic expansivity and the cubic expansivity of the container (

\gamma_r = \gamma_a + \gamma_v

).

3

Determine the change in temperature

\Delta T = 100\text{ K}

Subtract the initial temperature from the final temperature:

110^\circ\text{C} - 10^\circ\text{C} = 100\text{ K}

.

4

Calculate the final density of the liquid at

110^\circ\text{C}

\rho_2 = 800\text{ kg m}^{-3}

Density varies inversely with volumetric expansion according to

\rho_2 = \frac{\rho_1}{1 + \gamma_r \Delta T}

.

Key Concept

Relationship between real expansivity, apparent expansivity, container expansion, and density change in fluids

Question 77Question →

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

Show answer & explanation

Answer: 30

Answer

The initial length of the steel rod is

30\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 78Question →

A pneumatic cylinder in a hydraulic brake mechanism contains $0.050\text{ m}^3$ of an ideal gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas is first compressed isothermally to a volume of $0.020\text{ m}^3$ , and then heated at constant volume until its pressure reaches $6.00 \times 10^5\text{ Pa}$ . What is the final temperature of the gas in degrees Celsius?

327^\circ\text{C}

54^\circ\text{C}

600^\circ\text{C}

1227^\circ\text{C}

Show answer & explanation

Answer: $327^\circ\text{C}$

Answer

The final temperature of the gas is

327^\circ\text{C}

.

First, converting the initial temperature gives

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

. For the isothermal compression, the pressure increases to

P_2 = P_1 \times \frac{V_1}{V_2} = 1.20 \times 10^5 \times \frac{0.050}{0.020} = 3.00 \times 10^5\text{ Pa}

. Next, during the constant volume heating step to

P_3 = 6.00 \times 10^5\text{ Pa}

, the pressure doubles, requiring the absolute temperature to double from

300\text{ K}

to

600\text{ K}

. Converting

600\text{ K}

back to Celsius yields

600 - 273 = 327^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin.

T_1 = 27 + 273 = 300\text{ K}

.

Gas laws strictly require thermodynamic temperatures expressed in Kelvin.

2

Apply Boyle's Law (

P_1 V_1 = P_2 V_2

) to determine the intermediate pressure

P_2

after isothermal compression at constant temperature

T_2 = 300\text{ K}

.

P_2 = \frac{P_1 V_1}{V_2} = \frac{(1.20 \times 10^5\text{ Pa})(0.050\text{ m}^3)}{0.020\text{ m}^3} = 3.00 \times 10^5\text{ Pa}

.

During an isothermal process, the product of pressure and volume remains constant.

3

Apply Pressure Law (Gay-Lussac's Law) for the second stage, where volume remains constant (

V_3 = V_2 = 0.020\text{ m}^3

) while pressure increases from

P_2 = 3.00 \times 10^5\text{ Pa}

to

P_3 = 6.00 \times 10^5\text{ Pa}

.

\frac{P_2}{T_2} = \frac{P_3}{T_3} \implies T_3 = T_2 \times \frac{P_3}{P_2} = 300\text{ K} \times \frac{6.00 \times 10^5\text{ Pa}}{3.00 \times 10^5\text{ Pa}} = 600\text{ K}

.

At constant volume, pressure is directly proportional to absolute temperature.

4

Convert the final temperature

T_3

back to degrees Celsius.

t_3 = 600 - 273 = 327^\circ\text{C}

.

The question explicitly requests the final temperature in degrees Celsius.

Key Concept

Multi-stage gas processes combining Boyle's Law and Pressure Law using the Ideal Gas Equation

Question 79Question →

Match each kinetic theory parameter of an ideal gas on the left with its corresponding microscopic physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Temperature corresponds to the measure of average translational kinetic energy; Gas pressure corresponds to the rate of momentum transfer per unit area from wall collisions; Root-mean-square speed corresponds to the square root of the mean of squared speeds.

Temperature is directly linked to the average kinetic energy of molecules, gas pressure arises from wall collisions delivering impulse per unit area, and r.m.s. speed is the square root of mean square velocity.

Step-by-Step Solution

1

Identify the microscopic origin of temperature.

Temperature represents the average translational kinetic energy of gas molecules.

From the kinetic theory equation

\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T

, absolute temperature directly measures molecular kinetic energy.

2

Identify the microscopic origin of pressure.

Pressure is caused by molecular collisions with container walls.

Each collision transfers momentum to the wall; force is the time rate of momentum change, and force per unit area defines pressure.

3

Identify the definition of root-mean-square speed.

r.m.s. speed is the square root of the average of squared molecular speeds.

It accounts for the statistical distribution of molecular velocities in a gas sample.

Key Concept

Microscopic interpretation of macroscopic gas properties via Kinetic Theory of Matter

Question 80Question →

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

1.8 \times 10^{-5}\text{ K}^{-1}

3.6 \times 10^{-5}\text{ K}^{-1}

5.4 \times 10^{-5}\text{ K}^{-1}

1.08 \times 10^{-4}\text{ K}^{-1}

Show answer & explanation

Answer: $5.4 \times 10^{-5}\text{ K}^{-1}$

Answer

5.4 \times 10^{-5}\text{ K}^{-1}

The correct answer is derived by first finding the area expansivity

\beta = \frac{\Delta A}{A_0 \Delta T} = 3.6 \times 10^{-5}\text{ K}^{-1}

. Since

\beta = 2\alpha

, the linear expansivity

\alpha = 1.8 \times 10^{-5}\text{ K}^{-1}

. The cubical expansivity is

\gamma = 3\alpha = 5.4 \times 10^{-5}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change

\Delta T

\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

\beta

\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

\Delta A = A_0 \beta \Delta T

.

3

Calculate the coefficient of linear expansivity

\alpha

\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 (

\beta = 2\alpha

).

4

Calculate the coefficient of cubical expansivity

\gamma

\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 (

\gamma = 3\alpha

).

Key Concept

Relationship between linear (

\alpha

), area (

\beta

), and volume (

\gamma

) expansivities:

\beta = 2\alpha

and

\gamma = 3\alpha

.

Estimated Time:1m 30s

Thermal Physics

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