Thermal Physics Practice Questions — JAMB UTME — Page 5

Question 81Question →

A glass flask with an internal volume of $400\text{ cm}^3$ is filled to the brim with a liquid at $25^\circ\text{C}$ . The system is heated to $75^\circ\text{C}$ , causing $9.0\text{ cm}^3$ of liquid to overflow. Given that the linear expansivity of glass is $1.0 \times 10^{-5}\text{ K}^{-1}$ , what is the real cubic expansivity of the liquid in units of $10^{-4}\text{ K}^{-1}$ ?

Show answer & explanation

Answer: 4.8

Answer

The real cubic expansivity of the liquid is

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

, which equals

4.8

in units of

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

.

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

\gamma_r = \gamma_a + \gamma_v

). Calculating the apparent cubic expansivity yields

\gamma_a = \frac{9.0}{400 \times 50} = 4.5 \times 10^{-4}\text{ K}^{-1}

. The cubic expansivity of the glass container is

\gamma_v = 3 \times 1.0 \times 10^{-5} = 0.3 \times 10^{-4}\text{ K}^{-1}

. Adding these values gives a real cubic expansivity of

\gamma_r = 4.8 \times 10^{-4}\text{ K}^{-1}

, which equals

4.8

in units of

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

.

Step-by-Step Solution

1

Calculate the temperature change

\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50\text{ K}

Thermal expansion depends directly on the change in temperature.

2

Determine the apparent cubic expansivity of the liquid

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{9.0\text{ cm}^3}{400\text{ cm}^3 \times 50\text{ K}} = 4.5 \times 10^{-4}\text{ K}^{-1}

Apparent cubic expansivity is determined from the volume of liquid that overflows relative to the initial volume and temperature increase.

3

Calculate the cubic expansivity of the glass container

\gamma_v = 3\alpha = 3 \times 1.0 \times 10^{-5}\text{ K}^{-1} = 0.3 \times 10^{-4}\text{ K}^{-1}

The volumetric expansivity of an isotropic solid container is three times its linear expansivity.

4

Compute the real cubic expansivity of the liquid

\gamma_r = \gamma_a + \gamma_v = 4.5 \times 10^{-4}\text{ K}^{-1} + 0.3 \times 10^{-4}\text{ K}^{-1} = 4.8 \times 10^{-4}\text{ K}^{-1}

Real expansivity accounts for both the observed apparent expansion of the liquid and the expansion of the vessel containing it.

Key Concept

Thermal Expansion of Liquids and Anomalous Expansion of Water

Question 82Question →

A cylindrical metal rivet has a diameter of $2.50\text{ cm}$ at a room temperature of $25^\circ\text{C}$ . It needs to be inserted into a hole of diameter $2.49\text{ cm}$ 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 $2.0 \times 10^{-5}\text{ K}^{-1}$ ).

Show answer & explanation

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

\Delta d = d_0 \alpha \Delta T

. Substituting

\Delta d = -0.01\text{ cm}

,

d_0 = 2.50\text{ cm}

, and

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

gives

-0.01 = 2.50 \times (2.0 \times 10^{-5}) \times \Delta T

, leading to

\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 83Question →

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

Show answer & explanation

Answer: 140

Answer

140 °C

The required expansion in internal diameter is

\Delta d = 5.012\text{ cm} - 5.000\text{ cm} = 0.012\text{ cm}

. Using the linear expansion relation

\Delta d = d_0 \alpha \Delta T

, the required temperature change is

\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^\circ\text{C}

gives a final minimum temperature of

140^\circ\text{C}

.

Step-by-Step Solution

1

Determine the required increase in internal diameter (

\Delta d

) of the brass sleeve

\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

\Delta d = d_0 \alpha \Delta T

to find the temperature rise

\Delta T

\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

\alpha

.

3

Calculate the final temperature

T_2

T_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 84Question →

A sealed rigid canister contains a fixed mass of nitrogen gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until its temperature reaches $127^\circ\text{C}$ , what is the new pressure of the gas?

2.00 \times 10^5\text{ Pa}

7.06 \times 10^5\text{ Pa}

1.13 \times 10^5\text{ Pa}

3.19 \times 10^4\text{ Pa}

Show answer & explanation

Answer: $2.00 \times 10^5\text{ Pa}$

Answer

The new pressure of the gas is

2.00 \times 10^5\text{ Pa}

.

At constant volume, the pressure of a gas is directly proportional to its absolute temperature (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin gives

300\text{ K}

and

400\text{ K}

. Substituting these values yields

P_2 = 1.50 \times 10^5 \times (400/300) = 2.00 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Convert the initial and final temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 127 + 273 = 400\text{ K}

.

Gas laws require absolute temperature in Kelvin for proportional reasoning.

2

Apply Pressure Law (Gay-Lussac's Law) for constant volume.

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

.

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

3

Substitute the known values into the equation and solve for

P_2

.

P_2 = 1.50 \times 10^5\text{ Pa} \times \frac{400\text{ K}}{300\text{ K}} = 2.00 \times 10^5\text{ Pa}

.

Multiplying initial pressure by the temperature ratio gives the final pressure.

Key Concept

Pressure Law (Gay-Lussac's Law) states that at constant volume,

P \propto T

where

T

must be in Kelvin.

Question 85Question →

A sample of ideal gas in a syringe occupies a volume of $300\text{ cm}^3$ at a temperature of $27^\circ\text{C}$ . If the pressure of the gas remains constant, what is its volume in $\text{cm}^3$ when the temperature is raised to $127^\circ\text{C}$ ?

Show answer & explanation

Answer: 400

Answer

400 cm³

According to Charles's Law, at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature (

V \propto T

). Converting the given temperatures to kelvins yields

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Applying

\frac{V_1}{T_1} = \frac{V_2}{T_2}

gives

V_2 = \frac{300 \times 400}{300} = 400\text{ cm}^3

.

Step-by-Step Solution

1

Convert the initial and final temperatures from degrees Celsius to kelvins.

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

and

T_2 = 127 + 273 = 400\text{ K}

Gas law calculations require absolute temperature measured on the kelvin scale.

2

Apply Charles's Law, which relates volume and absolute temperature at constant pressure.

\frac{V_1}{T_1} = \frac{V_2}{T_2}

The pressure of the gas is maintained constant.

3

Substitute the values into the equation and solve for the final volume

V_2

.

\frac{300\text{ cm}^3}{300\text{ K}} = \frac{V_2}{400\text{ K}} \implies V_2 = 400\text{ cm}^3

Cross-multiplying gives

V_2 = \frac{300 \times 400}{300} = 400\text{ cm}^3

.

Key Concept

Charles's Law

Question 86Question →

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

3.0\text{ cm}^3

1.0\text{ cm}^3

2.0\text{ cm}^3

4.2\text{ cm}^3

Show answer & explanation

Answer: $3.0\text{ cm}^3$

Answer

The increase in the volume of the sphere is

3.0\text{ cm}^3

.

The value of

3.0\text{ cm}^3

is correct because the volume expansion requires the cubical expansivity

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

. Multiplying this coefficient by the initial volume (

1000\text{ cm}^3

) and the temperature change (

50\text{ K}

) yields

\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 (

\Delta T

).

\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 (

\gamma

) from linear expansivity (

\alpha

).

\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 (

\Delta V

).

\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 (

\gamma = 3\alpha

) and application of the volume expansion formula.

Estimated Time:1m 30s

Question 87Question →

A rigid solid object with negligible thermal expansivity is completely submerged in a container of water initially at $10^\circ\text{C}$ . The water is then uniformly cooled down to $0^\circ\text{C}$ . How does the upthrust (buoyant force) exerted by the water on the submerged object vary during this cooling process?

It increases to a maximum value at

4^\circ\text{C}

and then decreases as the temperature falls to

0^\circ\text{C}

.

It decreases continuously from

10^\circ\text{C}

down to

0^\circ\text{C}

as the liquid cools.

It remains constant throughout the cooling process because the volume of the submerged object does not change.

It decreases to a minimum value at

4^\circ\text{C}

and then increases as the temperature reaches

0^\circ\text{C}

.

Show answer & explanation

Answer: It increases to a maximum value at $4^\circ\text{C}$ and then decreases as the temperature falls to $0^\circ\text{C}$ .

Answer

The upthrust increases to a maximum value at

4^\circ\text{C}

and then decreases as cooling continues to

0^\circ\text{C}

.

According to Archimedes' principle, upthrust is given by

U = \rho_{\text{fluid}} V g

. For a submerged object of constant volume, upthrust is directly proportional to the density of water. As water cools from

10^\circ\text{C}

to

4^\circ\text{C}

, it contracts normally, increasing its density to a maximum at

4^\circ\text{C}

(

1000\text{ kg m}^{-3}

). Upon further cooling from

4^\circ\text{C}

to

0^\circ\text{C}

, water undergoes anomalous expansion, increasing in volume and decreasing in density. Consequently, the upthrust reaches a maximum at

4^\circ\text{C}

and decreases down to

0^\circ\text{C}

.

Step-by-Step Solution

1

Identify the expression for upthrust (buoyant force)

U = \rho_{\text{water}} \cdot V_{\text{object}} \cdot g

By Archimedes' principle, upthrust depends directly on the density of the displaced liquid

\rho_{\text{water}}

when the volume

V_{\text{object}}

and gravitational acceleration

g

are constant.

2

Analyze the density behavior of water from

10^\circ\text{C}

down to

4^\circ\text{C}

Density increases, reaching its maximum value

\rho_{\text{max}} = 1000\text{ kg m}^{-3}

at

4^\circ\text{C}

.

Between

10^\circ\text{C}

and

4^\circ\text{C}

, water contracts normally upon cooling.

3

Analyze the density behavior of water from

4^\circ\text{C}

down to

0^\circ\text{C}

Density decreases as temperature drops from

4^\circ\text{C}

to

0^\circ\text{C}

.

Water exhibits anomalous expansion between

4^\circ\text{C}

and

0^\circ\text{C}

, expanding in volume and consequently decreasing in density.

4

Relate density variation directly to upthrust variation

Upthrust increases from

10^\circ\text{C}

up to a peak at

4^\circ\text{C}

, then decreases from

4^\circ\text{C}

to

0^\circ\text{C}

.

Since upthrust is directly proportional to density, its value mirrors the density profile of water.

Key Concept

Anomalous expansion of water and its effect on density and upthrust

Question 88Question →

A block of mass $0.5\text{ kg}$ absorbs $4500\text{ J}$ of heat energy as its temperature rises from $25^\circ\text{C}$ to $45^\circ\text{C}$ . What is the heat capacity of the block?

450\text{ J K}^{-1}

225\text{ J K}^{-1}

112.5\text{ J K}^{-1}

225\text{ J kg}^{-1}\text{ K}^{-1}

Show answer & explanation

Answer: $225\text{ J K}^{-1}$

Answer

225\text{ J K}^{-1}

Heat capacity (

C

) is defined as thermal energy absorbed divided by temperature change:

C = \frac{Q}{\Delta T} = \frac{4500\text{ J}}{45\text{ K} - 25\text{ K}} = 225\text{ J K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

).

\Delta T = 45^\circ\text{C} - 25^\circ\text{C} = 20^\circ\text{C} = 20\text{ K}

Heat capacity depends on the temperature change in Kelvin or degrees Celsius.

2

Apply the heat capacity formula

C = \frac{Q}{\Delta T}

.

C = \frac{4500\text{ J}}{20\text{ K}} = 225\text{ J K}^{-1}

Heat capacity (

C

) measures the heat required to raise the temperature of the entire body by

1\text{ K}

, regardless of mass.

Key Concept

Heat capacity (

C

) represents the energy required to change an entire object's temperature by one kelvin (

C = \frac{Q}{\Delta T}

), whereas specific heat capacity (

c

) is per unit mass (

c = \frac{Q}{m\Delta T}

).

Estimated Time:1m 0s

Question 89Question →

An industrial compression chamber fitted with a frictionless piston contains $0.25\text{ m}^3$ of an ideal gas at an initial pressure of $1.6 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas expands while being heated to a final temperature of $87^\circ\text{C}$ until its pressure drops to $8.0 \times 10^4\text{ Pa}$ . What is the final volume occupied by the gas?

0.15\text{ m}^3

0.42\text{ m}^3

0.60\text{ m}^3

1.61\text{ m}^3

Show answer & explanation

Answer: $0.60\text{ m}^3$

Answer

0.60\text{ m}^3

The combined gas law states that

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

. Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Rearranging for

V_2

gives

V_2 = 0.25 \times \left(\frac{1.6 \times 10^5}{8.0 \times 10^4}\right) \times \left(\frac{360}{300}\right) = 0.25 \times 2.0 \times 1.2 = 0.60\text{ m}^3

.

Step-by-Step Solution

1

Convert all given temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 87 + 273 = 360\text{ K}

.

Gas laws require thermodynamic (absolute) temperature in Kelvin.

2

State the combined gas law relating initial and final states.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The mass of gas is fixed while pressure, volume, and temperature all change.

3

Rearrange the equation to solve for the final volume

V_2

.

V_2 = V_1 \times \left(\frac{P_1}{P_2}\right) \times \left(\frac{T_2}{T_1}\right)

Isolating the target unknown variable.

4

Substitute the given numerical values into the rearranged equation.

V_2 = 0.25\text{ m}^3 \times \left(\frac{1.6 \times 10^5\text{ Pa}}{8.0 \times 10^4\text{ Pa}}\right) \times \left(\frac{360\text{ K}}{300\text{ K}}\right) = 0.25 \times 2.0 \times 1.2 = 0.60\text{ m}^3

.

Simplifying the ratios yields the correct final volume.

Key Concept

Combined Gas Law

Question 90Question →

The specific heat capacity of a gas undergoing an isothermal expansion is zero because the temperature of the gas does not change during the process.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. During an isothermal process, the specific heat capacity of a gas is infinitely large (

\infty

), not zero.

The statement is false because the specific heat capacity

c = \frac{Q}{m \Delta T}

of a gas during an isothermal process (

\Delta T = 0, Q \neq 0

) is infinitely large (

\infty

). A specific heat capacity of zero occurs during an adiabatic process where no thermal energy enters or leaves the system (

Q = 0

) while temperature changes.

Step-by-Step Solution

1

Apply the fundamental defining equation for specific heat capacity.

c = \frac{Q}{m \Delta T}

, where

Q

is the quantity of heat transferred,

m

is the mass, and

\Delta T

is the change in temperature.

Specific heat capacity measures the amount of heat required per unit mass to produce a unit temperature change under a defined thermodynamic process.

2

Identify the thermodynamic boundary conditions for an isothermal expansion.

The temperature remains constant throughout the expansion, so

\Delta T = 0

, but heat energy

Q > 0

must be supplied to offset the work done by the expanding gas.

According to the First Law of Thermodynamics,

\Delta U = Q - W

. For an ideal gas undergoing an isothermal process,

\Delta U = 0

, which mandates

Q = W \neq 0

.

3

Evaluate the limit of

c

as

\Delta T \to 0

for a non-zero heat input

Q

.

c = \frac{Q}{m \cdot 0} \to \infty

.

Dividing a finite non-zero quantity of heat by a zero temperature change yields an infinitely large heat capacity.

4

Contrast this result with the condition required for a specific heat capacity of zero.

For

c = 0

, the heat input must be zero (

Q = 0

) while

\Delta T \neq 0

, which defines an adiabatic process.

Zero heat capacity means temperature changes without any heat transfer.

Key Concept

Thermodynamic process dependence of specific heat capacity

Question 91Question →

A sample of an ideal gas has an average translational kinetic energy of $E_k$ per molecule at an initial temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until the average kinetic energy per molecule doubles to $2E_k$ , what is the final temperature of the gas in degrees Celsius?

54^\circ\text{C}

327^\circ\text{C}

600^\circ\text{C}

927^\circ\text{C}

Show answer & explanation

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

Answer

The final temperature of the gas is

327^\circ\text{C}

.

In the kinetic theory of matter, average translational kinetic energy per molecule is directly proportional to absolute temperature (

E_k \propto T

). Initial temperature

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

. Doubling the kinetic energy doubles the Kelvin temperature to

600\text{ K}

. Subtracting

273

yields

327^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin.

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

Kinetic theory calculations require thermodynamic temperature measured on the absolute Kelvin scale.

2

Apply the proportional relationship between average translational kinetic energy and temperature.

Since

E_k \propto T

, doubling

E_k

means

T_2 = 2 \times T_1 = 2 \times 300\text{ K} = 600\text{ K}

.

The average kinetic energy of gas molecules is directly proportional to the absolute temperature.

3

Convert the final temperature from Kelvin back to degrees Celsius.

t_2 = 600\text{ K} - 273 = 327^\circ\text{C}

The question asks for the final temperature specifically in degrees Celsius.

Key Concept

Average translational kinetic energy of an ideal gas molecule is directly proportional to its absolute temperature (

E_k = \frac{3}{2} k_B T

).

Estimated Time:1m 30s

Question 92Question →

A gas sample enclosed in a rigid container of fixed volume has a root-mean-square (r.m.s.) speed of $500\text{ m s}^{-1}$ at a temperature of $127^\circ\text{C}$ . If the gas is heated until its pressure is quadrupled, what is the new r.m.s. speed of the gas molecules in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 1000

Answer

1000

According to kinetic theory, the pressure of a fixed volume of gas is directly proportional to its absolute temperature (

P \propto T

), and the root-mean-square speed of its molecules is proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

). Quadrupling the pressure quadruples the absolute temperature in Kelvin from

400\text{ K}

to

1600\text{ K}

. Since the speed scales as

\sqrt{4} = 2

, the initial r.m.s. speed of

500\text{ m s}^{-1}

doubles to

1000\text{ m s}^{-1}

.

Step-by-Step Solution

1

Convert the initial temperature to absolute temperature (Kelvin)

T_1 = 127^\circ\text{C} + 273 = 400\text{ K}

Kinetic theory relationships and gas laws require temperature in absolute units (Kelvin).

2

Determine the new absolute temperature based on the pressure change at constant volume

T_2 = 4 \times T_1 = 1600\text{ K}

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

P \propto T

). Therefore, quadrupling the pressure quadruples the absolute temperature.

3

Calculate the new root-mean-square speed using the square root relationship

v_2 = v_1 \sqrt{\frac{T_2}{T_1}} = 500 \times \sqrt{4} = 1000\text{ m s}^{-1}

Root-mean-square speed is proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

).

Key Concept

Relationship between microscopic kinetic parameters (r.m.s. speed) and macroscopic state variables (pressure and absolute temperature)

Question 93Question →

A $0.020\text{ kg}$ sample of a liquid metal at its melting point of $500^\circ\text{C}$ solidifies completely as thermal energy is extracted from it at a constant rate of $10\text{ W}$ . If the specific latent heat of fusion of the metal is $2.0 \times 10^4\text{ J kg}^{-1}$ , how long does the solidification process take?

40\text{ s}

400\text{ s}

2000\text{ s}

20000\text{ s}

Show answer & explanation

Answer: $40\text{ s}$

Answer

The solidification process takes

40\text{ s}

.

The thermal energy released when a substance solidifies at its melting point is given by

Q = m L_f

. Substituting

m = 0.020\text{ kg}

and

L_f = 2.0 \times 10^4\text{ J kg}^{-1}

gives

Q = 400\text{ J}

. Dividing this energy by the constant rate of heat removal (

10\text{ W}

) yields

t = \frac{400\text{ J}}{10\text{ W}} = 40\text{ s}

.

Step-by-Step Solution

1

Calculate the total thermal energy (

Q

) released during phase change at constant temperature

Q = m L_f = 0.020\text{ kg} \times 2.0 \times 10^4\text{ J kg}^{-1} = 400\text{ J}

Phase change occurs at a constant temperature, so thermal energy depends solely on mass and specific latent heat of fusion.

2

Calculate time (

t

) required using the power rate (

P

)

t = \frac{Q}{P} = \frac{400\text{ J}}{10\text{ W}} = 40\text{ s}

Power is defined as energy transferred per unit time (

P = Q / t

).

Key Concept

Latent Heat of Fusion and Energy Balance

Question 94Question →

A rigid scuba diving cylinder contains a fixed mass of compressed air at an initial absolute pressure of $2.00 \times 10^7\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The cylinder is left under direct sunlight on a boat deck, causing its temperature to rise to $77^\circ\text{C}$ . Assuming the volume of the cylinder remains constant, what is the new pressure of the air inside the cylinder?

2.33 \times 10^7\text{ Pa}

5.70 \times 10^7\text{ Pa}

1.71 \times 10^7\text{ Pa}

5.13 \times 10^6\text{ Pa}

Show answer & explanation

Answer: $2.33 \times 10^7\text{ Pa}$

Answer

The final pressure of the air inside the cylinder is

2.33 \times 10^7\text{ Pa}

.

According to the Pressure Law, the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature (

P \propto T

). Converting the given temperatures yields

T_1 = 300\text{ K}

and

T_2 = 350\text{ K}

. Substituting these into

P_2 = P_1(T_2 / T_1)

gives

2.00 \times 10^7 \times (350 / 300) = 2.33 \times 10^7\text{ Pa}

.

Step-by-Step Solution

1

Convert given temperatures from Celsius to Kelvin.

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

, and

T_2 = 77^\circ\text{C} + 273 = 350\text{ K}

.

All gas laws strictly require absolute temperature measured in Kelvin.

2

State and rearrange the Pressure Law (Gay-Lussac's Law) for constant volume.

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

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

3

Substitute the known values into the equation to calculate the final pressure.

P_2 = 2.00 \times 10^7\text{ Pa} \times \frac{350\text{ K}}{300\text{ K}} \approx 2.33 \times 10^7\text{ Pa}

.

Evaluates the final pressure accurately after heating.

Key Concept

Pressure Law (Gay-Lussac's Law) for Ideal Gases

Question 95Question →

Match each physical quantity or concept from the kinetic theory of gases on the left with its corresponding microscopic description or mathematical relation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Root-mean-square speed corresponds to

\sqrt{\frac{3kT}{m}}

; Average translational kinetic energy per molecule corresponds to

\frac{3}{2}kT

; Gas pressure corresponds to

\frac{1}{3}\rho v_{\text{rms}}^2

; and Absolute temperature corresponds to the macroscopic measure proportional to mean translational kinetic energy.

Each kinetic theory quantity correctly matches its corresponding microscopic formula and definition derived from fundamental assumptions of ideal gas particle behavior.

Step-by-Step Solution

1

Analyze the microscopic derivation of root-mean-square speed

From kinetic theory,

\bar{E}_k = \frac{1}{2}m v_{\text{rms}}^2 = \frac{3}{2}kT

, which yields

v_{\text{rms}} = \sqrt{\frac{3kT}{m}}

.

This establishes the relationship between molecular speed, temperature, and mass.

2

Identify the relationship for average translational kinetic energy per molecule

The average translational kinetic energy per molecule is given directly by

\bar{E}_k = \frac{3}{2}kT

.

The mean kinetic energy per degree of freedom is

\frac{1}{2}kT

, summing to

\frac{3}{2}kT

for three translational dimensions.

3

Relate macroscopic gas pressure to microscopic particle collisions

Gas pressure is expressed as

P = \frac{1}{3}\rho v_{\text{rms}}^2

based on continuous elastic collisions of gas molecules with the container walls.

Pressure represents the average force exerted per unit area by molecular collisions.

4

Define absolute temperature in terms of molecular kinetic energy

Absolute temperature

T

is the macroscopic physical property directly proportional to the average kinetic energy of the molecules.

This provides the thermodynamic definition of temperature from kinetic theory.

Key Concept

Microscopic properties of ideal gas molecules and kinetic derivation of pressure and temperature

Question 96Question →

During daytime in coastal regions, land heats up faster than the sea. The air above the land expands, becomes less dense, and rises, allowing cooler air from the ocean to move inland to form a sea breeze. Which mode of heat transfer is primarily responsible for this bulk movement of fluid, and what physical property change drives it?

Convection, driven by a decrease in air density due to thermal expansion

Conduction, driven by energy transfer through stationary molecular collisions at the land-air boundary

Radiation, driven by direct absorption of solar electromagnetic waves by atmospheric air

Convection, driven by an increase in air density due to thermal expansion

Show answer & explanation

Answer: Convection, driven by a decrease in air density due to thermal expansion

Answer

Convection, driven by a decrease in air density due to thermal expansion

Convection is the mode of heat transfer in liquids and gases where heat is carried from one place to another by the actual bulk movement of the heated fluid. As air over the land absorbs heat, thermal expansion causes its volume to increase and its density to decrease. The lighter, warm air rises and is replaced by cooler, denser air from over the water, creating a sea breeze.

Step-by-Step Solution

1

Identify the mode of heat transfer involved in fluids moving in bulk currents

Heat transfer in fluids involving actual physical displacement/movement of matter is convection.

Conduction occurs without net motion of the medium, and radiation occurs via electromagnetic waves.

2

Analyze the physical driver of convective currents

When air above the land is heated, it expands (

ΔV > 0

), reducing its density (

ρ = m/V

).

Lower density air experiences a net upward buoyant force, creating a low-pressure area near the ground into which cooler, denser air flows.

Key Concept

Convection in fluids and buoyancy driven by thermal expansion

Question 97Question →

Match each heat transfer scenario on the left with its dominant microscopic mechanism or physical pathway on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heat transfer through a copper rod matches energy transport dominated by free electron movement; heat transfer across an evacuated space matches energy transport via electromagnetic waves; heat transfer throughout water heated from the bottom matches energy transport by bulk fluid movement driven by density changes; heat transfer through a porcelain ceramic plate matches energy transport restricted strictly to lattice vibrational waves.

Each heat transfer scenario correctly pairs with its governing physical mechanism: copper conducts heat via free electrons and lattice vibrations, an evacuated space allows thermal energy propagation only through electromagnetic radiation, heated water circulates via density-driven convection currents, and porcelain conducts heat slowly and exclusively via lattice vibrational waves.

Step-by-Step Solution

1

Analyze heat conduction pathways in metals versus non-metallic solids

Metals possess free electrons that diffuse rapidly to transfer kinetic energy along with lattice vibrations. Non-metallic insulators lack mobile free electrons, so thermal conduction occurs at a much slower rate exclusively via lattice vibrations.

Understanding the atomic-level distinction between metallic conductors and non-metallic insulators.

2

Evaluate heat transfer in a medium-free region (vacuum)

Conduction and convection both depend on molecular collisions or particle transport, whereas thermal radiation is an electromagnetic wave phenomenon requiring no material medium.

Identifying radiation as the sole mode capable of propagating across a vacuum.

3

Analyze thermal behavior in fluids heated from below

Thermal expansion reduces the fluid density at the bottom. Gravitational buoyancy forces push the less dense fluid upward while denser, cooler fluid sinks, forming convection currents.

Establishing buoyancy and density differentials as the driving forces of convection.

Key Concept

Microscopic mechanisms of heat conduction, convection, and radiation

Question 98Question →

Match each physical heat transfer scenario on the left with its underlying physical mechanism or governing property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heat propagation along a solid copper bar matches free electron diffusion and lattice vibrations. Vertical circulation of water in a vessel matches temperature-dependent density variations causing buoyant fluid motion. Thermal energy transport from the Sun to the Earth matches propagation of electromagnetic waves requiring no material medium. Minimization of heat transport by silvered glass walls matches reflection of infrared radiation by low-emissivity surfaces.

Each physical scenario strictly corresponds to its defining heat transfer process: conduction in metals operates via free electron diffusion and lattice vibration; convection in heated liquids is driven by density changes under gravity; radiation from the Sun traverses space via electromagnetic waves without a physical medium; and silvered thermos coatings prevent radiative transfer by reflecting infrared radiation due to low emissivity.

Step-by-Step Solution

1

Identify the primary mechanism of heat conduction in metals

Conduction in metals relies on both atomic lattice vibrations and the motion of free conduction electrons.

Solids maintain fixed positions, preventing bulk mass displacement, so heat transfers microscopically.

2

Analyze fluid movement under thermal expansion

Heating fluid decreases its local density, causing warm regions to experience upward buoyant forces.

This setup establishes free thermal convection, which requires both a fluid medium and a gravitational field.

3

Evaluate energy transfer through a vacuum

Energy moving through empty space propagates as thermal electromagnetic waves.

Radiation is the unique mode of heat transfer that functions without a physical medium.

4

Examine radiative reflection by low-emissivity coatings

Polished silver coating acts as a mirror to infrared rays, reflecting radiant energy.

Low emissivity directly reduces the rate of radiant heat emission and absorption.

Key Concept

Distinct mechanisms of conduction, convection, and thermal radiation

Estimated Time:1m 30s

Question 99Question →

A spherical black body of radius $r$ at an initial temperature of $27^\circ\text{C}$ emits thermal radiation at a rate of $W$ . If the radius of the sphere is doubled and its temperature is increased to $327^\circ\text{C}$ , what is the new rate of thermal radiation emitted by the sphere in terms of $W$ ?

64W

32W

16W

8W

Show answer & explanation

Answer: $64W$

Answer

The new rate of thermal radiation emitted by the sphere is

64W

.

According to the Stefan-Boltzmann law, the rate of energy radiation from a black body is given by

P = \sigma A T^4

. Doubling the radius of a sphere increases its surface area by a factor of

2^2 = 4

. Converting temperatures from Celsius to Kelvin gives

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

and

T_2 = 327 + 273 = 600\text{ K}

, showing that the absolute temperature doubles (

T_2 / T_1 = 2

). Raising this temperature ratio to the fourth power yields

2^4 = 16

. Combining the area factor of

4

and the temperature factor of

16

results in a total radiation rate increase of

4 \times 16 = 64

times the original rate

W

.

Step-by-Step Solution

1

Express initial radiation rate using Stefan-Boltzmann law and sphere surface area formula

The total power radiated by a black body is given by

P = \sigma A T^4

. For a sphere of radius

r

,

A_1 = 4\pi r^2

. Absolute temperature

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

. Thus,

W = \sigma (4\pi r^2) (300)^4

.

Stefan's law requires absolute temperature in Kelvin and total surface area of the radiator.

2

Determine the scaled surface area and absolute temperature for the final state

New radius

r_2 = 2r \implies A_2 = 4\pi (2r)^2 = 4 A_1

. New temperature

T_2 = 327 + 273 = 600\text{ K} = 2 T_1

.

Surface area of a sphere scales quadratically with radius, and temperatures must be converted to Kelvin.

3

Calculate the ratio of the new radiation rate to the initial radiation rate

\frac{P_2}{P_1} = \frac{A_2}{A_1} \times \left(\frac{T_2}{T_1}\right)^4 = 4 \times (2)^4 = 4 \times 16 = 64

.

Radiated power is directly proportional to surface area and to the fourth power of absolute temperature.

4

State the new power in terms of

W

P_2 = 64 W

.

Multiplying the initial rate

W

by the overall scaling factor of 64 gives the final answer.

Key Concept

Stefan-Boltzmann Law of Radiation (

P = \epsilon \sigma A T^4

)

Question 100Question →

The boiling point of a liquid decreases when the external atmospheric pressure acting on its surface is reduced.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Boiling occurs when the saturated vapour pressure of a liquid equals the external atmospheric pressure. Decreasing the external atmospheric pressure reduces the saturated vapour pressure threshold required for boiling, allowing the liquid to boil at a lower temperature.

Step-by-Step Solution

1

Identify the condition required for boiling to occur.

Boiling takes place when the saturated vapour pressure of the liquid equals the external atmospheric pressure.

This is the fundamental physical definition of boiling.

2

Analyze the effect of reducing external atmospheric pressure.

Lower external pressure means the required saturated vapour pressure is reached at a lower temperature.

Saturated vapour pressure increases with temperature, so a lower target pressure corresponds to a lower boiling temperature.

Key Concept

Dependence of Boiling Point on External Atmospheric Pressure

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