Thermal Physics Practice Questions — JAMB UTME — Page 6

Question 101Question →

A composite cylindrical bar consists of two uniform sections of equal length joined end-to-end. Section A has a radius of $2.0\text{ cm}$ and a thermal conductivity of $300\text{ W m}^{-1}\text{K}^{-1}$ . Section B has a radius of $4.0\text{ cm}$ and a thermal conductivity of $150\text{ W m}^{-1}\text{K}^{-1}$ . The outer end of Section A is maintained at a constant temperature of $120^\circ\text{C}$ , while the outer end of Section B is held at $0^\circ\text{C}$ . Assuming the curved surfaces of both sections are perfectly insulated and heat flow is steady, what is the temperature at the junction between the two sections in $^\circ\text{C}$ ?

Show answer & explanation

Answer: 40

Answer

The steady-state temperature at the junction between Section A and Section B is 40.0°C.

At steady state, the rate of heat conduction through Section A equals that through Section B. Because Section B has double the radius of Section A, its cross-sectional area is four times as large. Equating the heat flow rates gives 300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J, which simplifies directly to 120 - T_J = 2 T_J, yielding a junction temperature of 40.0°C.

Step-by-Step Solution

1

Determine the relationship between the cross-sectional areas of Section A and Section B.

The area ratio A_B / A_A = (r_B / r_A)² = (4.0 cm / 2.0 cm)² = 4.

The cross-sectional area of a cylinder is proportional to the square of its radius.

2

Write the steady-state heat flow equation for each section.

H_A = (k_A * A_A * (120 - T_J)) / L and H_B = (k_B * A_B * (T_J - 0)) / L.

According to Fourier's law of thermal conduction, the heat transfer rate through a uniform layer is proportional to thermal conductivity, cross-sectional area, and temperature difference, and inversely proportional to length.

3

Equate the heat transfer rates H_A and H_B and solve for the junction temperature T_J.

300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J => 300 * (120 - T_J) = 600 * T_J => 120 - T_J = 2 T_J => 3 T_J = 120 => T_J = 40.0°C.

At steady state with insulated sides, heat does not accumulate or escape, so the rate of heat conduction through both sections must be identical.

Key Concept

Steady-state thermal conduction through composite conductors with differing cross-sectional areas and thermal conductivities

Question 102Question →

A cylindrical copper rod of thermal conductivity $380\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , length $0.40\text{ m}$ , and uniform radius $2.0\text{ cm}$ is thermally insulated along its curved surface. One flat end is maintained at a temperature of $100^\circ\text{C}$ by steam, while the opposite end is kept in an ice bath at $0^\circ\text{C}$ . Assuming steady-state heat conduction, what is the rate of heat flow through the rod? (Take $\pi = 3.14$ )

119.3\text{ W}

19.1\text{ W}

1900.0\text{ W}

1193.2\text{ W}

Show answer & explanation

Answer: $119.3\text{ W}$

Answer

The rate of heat flow through the copper rod is

119.3\text{ W}

.

According to Fourier's law of heat conduction, the rate of thermal energy transfer

\frac{Q}{t}

through a material of thermal conductivity

k

, cross-sectional area

A

, and length

d

across temperature difference

\Delta T

is given by

\frac{Q}{t} = \frac{k A \Delta T}{d}

. Substituting

k = 380\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = \pi (0.02\text{ m})^2 = 1.256 \times 10^{-3}\text{ m}^2

,

\Delta T = 100\text{ K}

, and

d = 0.40\text{ m}

yields

119.3\text{ W}

.

Step-by-Step Solution

1

Convert the radius from centimeters to meters and calculate the cross-sectional area of the rod.

r = 2.0\text{ cm} = 0.02\text{ m}

.

A = \pi r^2 = 3.14 \times (0.02\text{ m})^2 = 1.256 \times 10^{-3}\text{ m}^2

.

Fourier's law requires the area in square meters (

m^2

).

2

Determine the temperature difference across the rod.

\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100\text{ K}

.

Heat transfer rate depends on the temperature gradient along the length.

3

Apply Fourier's law of thermal conduction:

\frac{Q}{t} = \frac{k A \Delta T}{d}

.

\frac{Q}{t} = \frac{380 \times 1.256 \times 10^{-3} \times 100}{0.40} = 119.32\text{ W} \approx 119.3\text{ W}

.

Substituting the thermal conductivity

k

, cross-sectional area

A

, temperature difference

\Delta T

, and length

d

gives the steady-state heat conduction rate.

Key Concept

Fourier's Law of Heat Conduction in solids:

\frac{Q}{t} = \frac{k A (T_{\text{hot}} - T_{\text{cold}})}{d}

Estimated Time:2m 0s

Question 103Question →

Two rectangular slabs of equal thickness $d$ are mounted in parallel between a hot reservoir at $100^\circ\text{C}$ and a cold reservoir at $20^\circ\text{C}$ . Slab 1 has a thermal conductivity $k_1 = 300\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ and a cross-sectional area $A_1 = 4.0 \times 10^{-4}\text{ m}^2$ . Slab 2 has a thermal conductivity $k_2 = 100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ and a cross-sectional area $A_2 = 6.0 \times 10^{-4}\text{ m}^2$ . Assuming steady-state heat conduction and no lateral heat loss, what percentage of the total heat transferred per second between the reservoirs conducts through Slab 1?

66.7\%

33.3\%

50.0\%

40.0\%

Show answer & explanation

Answer: $66.7\%$

Answer

The percentage of the total heat transferred per second conducted through Slab 1 is 66.7%.

Fourier's law gives the rate of heat conduction as

P = \frac{k A \Delta T}{d}

. Since the temperature gradient

\frac{\Delta T}{d}

is identical across both parallel slabs, the heat current through each slab is directly proportional to its

k A

product. For Slab 1,

k_1 A_1 = 300 \times 4.0 \times 10^{-4} = 0.12

, while for Slab 2,

k_2 A_2 = 100 \times 6.0 \times 10^{-4} = 0.06

. The total heat current is proportional to

0.12 + 0.06 = 0.18

. Thus, the fraction conducted through Slab 1 is

\frac{0.12}{0.18} = \frac{2}{3}

, which corresponds to

66.7\%

.

Step-by-Step Solution

1

Express Fourier's law of heat conduction for each slab in parallel.

Rate of heat flow

P_1 = \frac{k_1 A_1 \Delta T}{d}

and

P_2 = \frac{k_2 A_2 \Delta T}{d}

.

Both slabs experience the same temperature difference

\Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80\text{ K}

and have equal length

d

.

2

Calculate the effective conductance factors

k A

for both slabs.

k_1 A_1 = 300 \times (4.0 \times 10^{-4}) = 0.12\text{ W}\cdot\text{m}\cdot\text{K}^{-1}

and

k_2 A_2 = 100 \times (6.0 \times 10^{-4}) = 0.06\text{ W}\cdot\text{m}\cdot\text{K}^{-1}

.

Since

\Delta T / d

is identical for both parallel paths, the heat flow rate is directly proportional to

k A

.

3

Determine the total heat flow rate and the percentage carried by Slab 1.

P_{\text{total}} = P_1 + P_2 \propto 0.12 + 0.06 = 0.18

. Percentage through Slab 1

= \frac{0.12}{0.18} \times 100\% = 66.7\%

.

In parallel conduction, individual heat currents add to form the total heat current.

Key Concept

Parallel Thermal Conduction

Question 104Question →

Two solid cylindrical copper rods, $P$ and $Q$ , are maintained under identical temperature differences across their ends. Rod $P$ has twice the radius and half the length of Rod $Q$ . What is the ratio of the rate of heat conduction through Rod $P$ to that through Rod $Q$ ?

8 : 1

4 : 1

2 : 1

1 : 8

Show answer & explanation

Answer: $8 : 1$

Answer

The ratio of the rate of heat conduction through Rod

P

to that through Rod

Q

is

8 : 1

.

The rate of heat conduction is given by

\frac{Q}{t} = \frac{k \pi r^2 \Delta T}{L}

. For Rod

P

, substituting

r_P = 2 r_Q

and

L_P = 0.5 L_Q

gives

\frac{(2)^2}{0.5} = 8

times the rate of Rod

Q

. Thus, the ratio of heat conduction rate is

8 : 1

.

Step-by-Step Solution

1

Write the fundamental equation for the rate of thermal conduction through a uniform conductor.

\frac{Q}{t} = \frac{k A \Delta T}{L}

Heat conduction rate is proportional to thermal conductivity

k

, cross-sectional area

A

, temperature difference

\Delta T

, and inversely proportional to length

L

.

2

Express the cross-sectional area

A

in terms of radius

r

for a cylindrical rod.

A = \pi r^2 \implies \frac{Q}{t} = \frac{k \pi r^2 \Delta T}{L}

The cross-section of a cylindrical rod is a circle.

3

Set up the ratio of heat conduction rates for Rod

P

and Rod

Q

given

r_P = 2 r_Q

and

L_P = 0.5 L_Q

.

\frac{(Q/t)_P}{(Q/t)_Q} = \frac{r_P^2 / L_P}{r_Q^2 / L_Q} = \frac{(2 r_Q)^2 / (0.5 L_Q)}{r_Q^2 / L_Q} = \frac{4}{0.5} = 8

Material (

k

) and temperature difference (

\Delta T

) are identical for both rods and cancel out.

Key Concept

Thermal Conduction Rate Formula

Question 105Question →

Match each heat transfer process or physical phenomenon on the left with its underlying governing mechanism or quantitative relationship on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The steady-state rate of heat conduction through a uniform slab corresponds to being directly proportional to the temperature gradient. The radiant energy emitted per unit area by an ideal blackbody corresponds to being directly proportional to the fourth power of absolute temperature. Natural heat transport in fluids under gravity corresponds to being driven by buoyant forces resulting from density variations. The microscopic conduction mechanism in electrical insulators corresponds to propagation via quantized lattice vibrations (phonons).

Each heat transfer mechanism matches its fundamental law and microscopic process: conduction across a plane wall is governed by Fourier's law and proportional to the temperature gradient; thermal radiation from a blackbody obeys Stefan's law and scales with the fourth power of absolute temperature; natural convection in fluids requires gravity to drive density-based buoyant circulation; and thermal conduction in non-metallic insulators relies on atomic lattice vibrations (phonons) due to the absence of free electrons.

Step-by-Step Solution

1

Analyze conduction governing equation (Fourier's Law)

Heat current

\frac{Q}{t} = kA \frac{\Delta T}{d}

, showing that heat flow per unit area depends directly on the temperature gradient

\frac{\Delta T}{\Delta x}

.

Identify the quantitative relationship governing thermal conduction in solid materials.

2

Analyze radiation power equation (Stefan-Boltzmann Law)

Total power per unit area

P/A = \sigma T^4

, establishing fourth-power dependence on thermodynamic temperature

T

.

Identify the law governing thermal radiation emissions.

3

Examine natural convection mechanics

Thermal expansion leads to density differences

\Delta \rho

, causing buoyant forces under gravity to set up fluid circulation currents.

Identify the physical drive behind natural convection in fluids.

4

Examine microscopic heat transfer mechanisms in insulators

Insulators lack mobile valence electrons, leaving atomic lattice vibrations (phonons) as the sole mechanism for thermal energy transport.

Distinguish between electronic conduction in metals and lattice/phonon conduction in non-metals.

Key Concept

Physical Principles and Microscopic Mechanisms of Conduction, Convection, and Radiation

Question 106Question →

The specific heat capacity of a physical body increases proportionally with its mass, whereas its total heat capacity remains constant for any quantity of a given material.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Specific heat capacity is an intensive property independent of mass, whereas heat capacity is an extensive property that depends directly on mass.

The statement is false because specific heat capacity (

c

) is an intensive property that remains constant regardless of mass, while heat capacity (

C

) is an extensive property that increases directly with the mass of the object.

Step-by-Step Solution

1

Recall the mathematical definitions of heat capacity and specific heat capacity

Heat capacity

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

, whereas specific heat capacity

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

.

Establishing the functional dependence of both thermal quantities on mass.

2

Compare intensive and extensive classifications

Specific heat capacity

c

is normalized per unit mass (

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

) and is independent of mass. Heat capacity

C

(

\text{J K}^{-1}

) scales directly with mass

m

.

The statement falsely reverses which quantity depends on mass.

Key Concept

Distinguishing between extensive (heat capacity) and intensive (specific heat capacity) thermal properties

Estimated Time:1m 0s

Question 107Question →

A sample of gas enclosed in a vessel has a density of $0.90\text{ kg/m}^3$ and exerts a pressure of $3.0 \times 10^5\text{ N/m}^2$ on the walls of the vessel. Based on the kinetic theory of gases, what is the root-mean-square (r.m.s.) speed of the gas molecules in $\text{m/s}$ ?

Show answer & explanation

Answer: 1000

Answer

The root-mean-square speed of the gas molecules is

1000\text{ m/s}

.

By applying the kinetic theory formula

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

, rearranging gives

v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

. Substituting

P = 3.0 \times 10^5\text{ N/m}^2

and

\rho = 0.90\text{ kg/m}^3

results in

v_{\text{rms}} = \sqrt{\frac{9.0 \times 10^5}{0.90}} = \sqrt{1.0 \times 10^6} = 1000\text{ m/s}

.

Step-by-Step Solution

1

Identify the kinetic theory equation relating gas pressure, density, and microscopic molecular speed.

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

According to the kinetic theory of gases, the macroscopic pressure exerted by gas molecules colliding with container walls is proportional to the gas density and the square of their r.m.s. speed.

2

Make

v_{\text{rms}}

the subject of the formula.

v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

Multiplying both sides by

3

and dividing by density

\rho

isolates

v_{\text{rms}}^2

, taking the square root yields

v_{\text{rms}}

.

3

Substitute the given numerical values into the equation.

v_{\text{rms}} = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.90}} = \sqrt{1,000,000} = 1000\text{ m/s}

Performing the division yields

1.0 \times 10^6\text{ m}^2/\text{s}^2

, whose square root gives the speed in

\text{m/s}

.

Key Concept

Kinetic Theory Pressure Equation relating macroscopic pressure and density to microscopic root-mean-square velocity (

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

).

Question 108Question →

Match each thermometer type on the left with its corresponding physical thermometric property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Constant-volume gas thermometer matches variation of gas pressure at constant volume; Thermocouple matches variation of electromotive force (e.m.f.); Platinum resistance thermometer matches variation of electrical resistance; Liquid-in-glass thermometer matches variation of liquid column length/volume.

Each thermometer is paired with its corresponding thermometric property: constant-volume gas thermometer relies on gas pressure variation, thermocouple relies on thermoelectric e.m.f. variation, resistance thermometer relies on electrical resistance variation, and liquid-in-glass thermometer relies on liquid thermal expansion.

Step-by-Step Solution

1

Identify the thermometric property for a constant-volume gas thermometer.

Gas pressure at constant volume.

Pressure varies linearly with temperature for an ideal gas at constant volume.

2

Identify the thermometric property for a thermocouple.

Electromotive force (e.m.f.).

The thermoelectric effect generates an e.m.f. proportional to the temperature difference between two junctions.

3

Identify the thermometric property for a platinum resistance thermometer.

Electrical resistance.

The electrical resistance of platinum increases continuously and predictably with temperature.

4

Identify the thermometric property for a liquid-in-glass thermometer.

Liquid column length/volume.

The liquid expands linearly along the capillary stem as temperature rises.

Key Concept

Thermometric Properties of Thermometers

Question 109Question →

A thermocouple thermometer produces an electromotive force (e.m.f.) of $2.0\,\text{mV}$ at the ice point ( $0^\circ\text{C}$ ) and $18.0\,\text{mV}$ at the steam point ( $100^\circ\text{C}$ ). When placed in a liquid bath, the recorded e.m.f. is $14.0\,\text{mV}$ . What is the temperature of the liquid bath on the Celsius scale?

66.7^\circ\text{C}

70.0^\circ\text{C}

75.0^\circ\text{C}

87.5^\circ\text{C}

Show answer & explanation

Answer: $75.0^\circ\text{C}$

Answer

75.0^\circ\text{C}

The temperature on a linear scale is proportional to the fraction of the interval traversed between the fixed points. Subtracting the baseline reading of

2.0\,\text{mV}

gives an effective increase of

12.0\,\text{mV}

out of a total range of

16.0\,\text{mV}

. Multiplying this fraction (

0.75

) by

100^\circ\text{C}

yields

75.0^\circ\text{C}

.

Step-by-Step Solution

1

Identify the given thermometric values for the lower fixed point, upper fixed point, and unknown temperature reading.

E_0 = 2.0\,\text{mV}

,

E_{100} = 18.0\,\text{mV}

, and

E_\theta = 14.0\,\text{mV}

.

Linear temperature scales relate the change in thermometric property proportionally to temperature changes.

2

Apply the standard linear interpolation formula for a Celsius temperature scale.

\theta = \frac{E_\theta - E_0}{E_{100} - E_0} \times 100^\circ\text{C}

This accounts for the baseline reading at the ice point (

0^\circ\text{C}

) and normalizes it over the

100^\circ\text{C}

fundamental interval.

3

Substitute the known values and evaluate the expression.

\theta = \frac{14.0 - 2.0}{18.0 - 2.0} \times 100 = \frac{12.0}{16.0} \times 100 = 0.75 \times 100 = 75.0^\circ\text{C}

.

Carrying out the subtraction yields a proportional change of

\frac{3}{4}

of the full fundamental interval.

Key Concept

Temperature Scale Calibration and Thermocouple Interpolation

Estimated Time:1m 30s

Question 110Question →

A constant-volume gas thermometer registers a pressure of $60\,\text{kPa}$ at the ice point ( $0^\circ\text{C}$ ) and $84\,\text{kPa}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature in degrees Celsius when the pressure registered by the thermometer is $72\,\text{kPa}$ ?

Show answer & explanation

Answer: 50

Answer

The temperature corresponding to a pressure reading of

72\,\text{kPa}

is

50^\circ\text{C}

.

Using the linear relation for a constant-volume gas thermometer:

T = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

. Substituting

P_T = 72\,\text{kPa}

,

P_0 = 60\,\text{kPa}

, and

P_{100} = 84\,\text{kPa}

yields

T = \frac{72 - 60}{84 - 60} \times 100 = \frac{12}{24} \times 100 = 50^\circ\text{C}

.

Step-by-Step Solution

1

Identify the thermometric property values at the fixed points and target state.

P_0 = 60\,\text{kPa}

,

P_{100} = 84\,\text{kPa}

, and

P_T = 72\,\text{kPa}

.

These represent the pressure values corresponding to

0^\circ\text{C}

,

100^\circ\text{C}

, and the unknown temperature

T

respectively.

2

Set up the linear scale conversion equation.

T = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}

Temperature changes linearly with the thermometric property (gas pressure at constant volume).

3

Calculate the numerical value.

T = \frac{12}{24} \times 100 = 50^\circ\text{C}

Simplifying the fraction

\frac{12}{24} = 0.5

and multiplying by

100

gives

50

.

Key Concept

Temperature measurement using constant-volume gas pressure as a thermometric property

Question 111Question →

If a sample of unsaturated air with a relative humidity of $40\%$ at $30^\circ\text{C}$ is compressed isothermally to one-third of its original volume, the final relative humidity of the air will be $100\%$ .

True

False

Show answer & explanation

Answer: True

Answer

True. The relative humidity cannot exceed 100% because water vapour condenses into liquid once its partial pressure reaches the saturated vapour pressure at that temperature.

The statement is true because relative humidity is the ratio of actual partial vapour pressure to saturated vapour pressure at a specific temperature. Although isothermal compression to one-third volume triples the partial pressure of water vapour (giving a theoretical 120%), partial vapour pressure is bounded by the saturated vapour pressure. When the relative humidity reaches 100%, condensation occurs, keeping the relative humidity at exactly 100%.

Step-by-Step Solution

1

Apply Boyle's law to calculate the theoretical partial vapour pressure after isothermal compression.

Since volume is reduced to

\frac{1}{3}V_1

at constant temperature, the partial pressure of water vapour would increase by a factor of 3 (

p_2 = 3p_1

).

For an unsaturated vapour at constant temperature, partial pressure is inversely proportional to volume.

2

Determine the theoretical relative humidity from the pressure ratio.

\text{Theoretical R.H.} = 3 \times 40\% = 120\%.

Relative humidity is defined as

\text{R.H.} = \frac{p}{p_s} \times 100\%

, where

p

is the partial pressure and

p_s

is the saturated vapour pressure.

3

Apply the physical limit imposed by saturated vapour pressure.

The actual relative humidity cannot exceed

100\%

; excess vapour condenses.

Saturated vapour pressure

p_s

represents the maximum possible partial pressure of water vapour in air at a given temperature.

Key Concept

Saturated Vapour Pressure Limit and Relative Humidity under Isothermal Compression

Question 112Question →

A composite solid consists of a metal block $A$ of mass $2.0\text{ kg}$ and a metal block $B$ of mass $1.0\text{ kg}$ joined in thermal contact. Block $A$ has a specific heat capacity of $300\text{ J kg}^{-1}\text{K}^{-1}$ . When the composite solid absorbs $18\text{ kJ}$ of thermal energy, its overall temperature increases by $15\text{ K}$ . Assuming no heat loss to the surroundings, what is the specific heat capacity of metal block $B$ ?

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

1200\text{ J kg}^{-1}\text{K}^{-1}

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

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

Show answer & explanation

Answer: $600\text{ J kg}^{-1}\text{K}^{-1}$

Answer

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

The total heat energy supplied (

18\text{ kJ} = 18000\text{ J}

) causes a temperature rise of

15\text{ K}

, giving a total heat capacity

C_{\text{total}} = 1200\text{ J K}^{-1}

for the composite solid. Subtracting block

A

's heat capacity (

C_A = 2.0\text{ kg} \times 300\text{ J kg}^{-1}\text{K}^{-1} = 600\text{ J K}^{-1}

) leaves a heat capacity of

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

for block

B

. Dividing block

B

's heat capacity by its mass (

1.0\text{ kg}

) yields its specific heat capacity of

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

.

Step-by-Step Solution

1

Calculate the total heat capacity (

C_{\text{total}}

) of the composite system.

C_{\text{total}} = \frac{Q}{\Delta T} = \frac{18000\text{ J}}{15\text{ K}} = 1200\text{ J K}^{-1}

Heat capacity is defined as the total heat energy absorbed per unit change in temperature.

2

Determine the heat capacity (

C_A

) of metal block

A

.

C_A = m_A \times c_A = 2.0\text{ kg} \times 300\text{ J kg}^{-1}\text{K}^{-1} = 600\text{ J K}^{-1}

Heat capacity is the product of mass and specific heat capacity.

3

Determine the heat capacity (

C_B

) of metal block

B

.

C_B = C_{\text{total}} - C_A = 1200\text{ J K}^{-1} - 600\text{ J K}^{-1} = 600\text{ J K}^{-1}

Total heat capacity of a composite body is the sum of the individual heat capacities of its components.

4

Calculate the specific heat capacity (

c_B

) of metal block

B

.

c_B = \frac{C_B}{m_B} = \frac{600\text{ J K}^{-1}}{1.0\text{ kg}} = 600\text{ J kg}^{-1}\text{K}^{-1}

Specific heat capacity is the heat capacity per unit mass of that specific substance.

Key Concept

Additivity of Heat Capacity in Composite Systems (

C_{\text{total}} = m_A c_A + m_B c_B

)

Question 113Question →

A mass of air at $30^\circ\text{C}$ has a relative humidity of $50\%$ . The saturated vapour pressure of water is $30\text{ mmHg}$ at $30^\circ\text{C}$ and $9\text{ mmHg}$ at $10^\circ\text{C}$ . If the temperature of the air is lowered to $10^\circ\text{C}$ , what percentage of the initial water vapour condenses out?

$40\%

$60\%

$70\%

$30\%

Show answer & explanation

Answer: $40\%

Answer

The percentage of initial water vapour that condenses out when cooled to

10^\circ\text{C}

is

40\%

.

The correct answer is

40\%

. Initially, the air contains water vapour exerting a partial pressure of

0.50 \times 30\text{ mmHg} = 15\text{ mmHg}

. Upon cooling to

10^\circ\text{C}

, the air becomes saturated at

9\text{ mmHg}

, causing

15\text{ mmHg} - 9\text{ mmHg} = 6\text{ mmHg}

worth of vapour to condense into liquid. The condensed amount as a fraction of the initial vapour is

6 / 15 = 0.40

, or

40\%

.

Step-by-Step Solution

1

Calculate the initial partial vapour pressure of water at

30^\circ\text{C}

.

\text{Partial Vapour Pressure} = \text{Relative Humidity} \times \text{SVP at } 30^\circ\text{C} = 0.50 \times 30\text{ mmHg} = 15\text{ mmHg}

.

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the amount of vapour pressure that must condense when cooled to

10^\circ\text{C}

.

\text{Vapour pressure condensed} = 15\text{ mmHg} - 9\text{ mmHg} = 6\text{ mmHg}

.

At

10^\circ\text{C}

, the air can hold at most its saturated vapour pressure of

9\text{ mmHg}

, so any excess vapour above

9\text{ mmHg}

condenses into liquid water.

3

Calculate the percentage of the initial water vapour that condenses out.

\text{Percentage condensed} = \left(\frac{6\text{ mmHg}}{15\text{ mmHg}}\right) \times 100\% = 40\%

.

The question asks for the fraction of the initial vapour originally present that leaves the gaseous state.

Key Concept

Relative Humidity and Dew Point Condensation

Estimated Time:1m 30s

Question 114Question →

A solid cylindrical metal rod of length $0.50\text{ m}$ and cross-sectional area $4.0 \times 10^{-4}\text{ m}^2$ is perfectly insulated along its lateral surface. One end of the rod is maintained at $100^\circ\text{C}$ in boiling water, while the other end is in contact with an ice block at $0^\circ\text{C}$ . If $0.072\text{ kg}$ of ice melts in $10\text{ minutes}$ due to thermal energy conducted through the rod, what is the thermal conductivity of the metal? (Take the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J}\cdot\text{kg}^{-1}$ )

630\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

7000\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

3.23\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

Show answer & explanation

Answer: $504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$

Answer

The thermal conductivity of the metal rod is

504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

The correct answer is derived by first finding the total heat absorbed during the phase change of ice using

Q = m L = 0.072 \times 3.36 \times 10^5 = 24,192\text{ J}

. Dividing by the time in seconds (

600\text{ s}

) gives a heat flow rate of

40.32\text{ W}

. Substituting this into the conduction formula

\frac{Q}{t} = \frac{k A \Delta T}{d}

gives

40.32 = \frac{k (4.0 \times 10^{-4})(100)}{0.50} = 0.08 k

, yielding

k = 504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

Step-by-Step Solution

1

Calculate the total heat energy

Q

required to melt

0.072\text{ kg}

of ice at

0^\circ\text{C}

.

Q = m L = 0.072\text{ kg} \times 3.36 \times 10^5\text{ J}\cdot\text{kg}^{-1} = 24,192\text{ J}

During melting at constant temperature, heat transfer is governed by the latent heat of fusion formula.

2

Convert the elapsed time into seconds and calculate the rate of heat transfer

\frac{Q}{t}

.

t = 10\text{ min} = 600\text{ s}

;

\frac{Q}{t} = \frac{24,192\text{ J}}{600\text{ s}} = 40.32\text{ W}

Thermal conductivity formulas require rate of heat transfer in Joules per second (Watts).

3

Apply Fourier's law of thermal conduction

\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d}

to solve for thermal conductivity

k

.

40.32 = \frac{k \times (4.0 \times 10^{-4}) \times (100 - 0)}{0.50} \implies 40.32 = 0.08 k \implies k = 504\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

Rearranging the steady-state thermal conduction equation yields

k = \frac{(Q/t) \cdot d}{A \cdot \Delta T}

.

Key Concept

Thermal Conduction Rate and Latent Heat of Fusion

Question 115Question →

Match each kinetic theory concept or microscopic property of an ideal gas on the left with its corresponding mathematical expression or derivation result on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair the momentum change per collision with

2 m v_x

, the single-molecule average wall force with

\frac{m v_x^2}{L}

, the kinetic energy density with

\frac{3}{2} P

, and the root-mean-square speed with

\sqrt{\frac{3 R T}{M}}

.

Each kinetic theory quantity is derived directly from fundamental principles of mechanics applied to gas particles. Elastic collision with a wall yields a momentum reversal of magnitude

2 m v_x

. Taking the round-trip collision frequency over length

L

yields an average force of

\frac{m v_x^2}{L}

. Linking microscopic kinetic energy density to pressure gives

\frac{E_k}{V} = \frac{3}{2} P

, and linking pressure to the ideal gas law for one mole yields

v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

.

Step-by-Step Solution

1

Analyze momentum transfer during elastic collision of a molecule with a wall.

Initial momentum along the x-axis is

p_i = m v_x

and final momentum after elastic reflection is

p_f = -m v_x

. The change in momentum is

\Delta p_x = p_f - p_i = -2 m v_x

, which has a magnitude of

2 m v_x

.

Elastic collision conserves kinetic energy and reverses velocity direction perpendicular to the wall.

2

Calculate the time rate of momentum transfer to determine average force.

The round-trip distance between opposite walls separated by length

L

is

2L

, so the time between collisions with the same wall is

\Delta t = \frac{2L}{v_x}

. The average force is

F_x = \frac{\Delta p}{\Delta t} = \frac{2 m v_x}{2L / v_x} = \frac{m v_x^2}{L}

.

Newton's second law expresses force as the average rate of change of momentum.

3

Relate total translational kinetic energy density to gas pressure.

From kinetic theory, gas pressure is given by

P = \frac{1}{3} \frac{N m}{V} v_{\text{rms}}^2

. Since total kinetic energy

E_k = \frac{1}{2} N m v_{\text{rms}}^2

, we can express pressure as

P = \frac{2}{3} \left(\frac{E_k}{V}\right)

. Rearranging gives energy density

\frac{E_k}{V} = \frac{3}{2} P

.

Translational kinetic energy density is directly proportional to pressure with a factor of 3/2.

4

Derive the formula for root-mean-square velocity from macroscopic and microscopic gas equations.

Substitute density

\rho = \frac{M}{V}

(where

M

is molar mass) into

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

, obtaining

P = \frac{M v_{\text{rms}}^2}{3 V}

. Since

P V = R T

for one mole of ideal gas,

R T = \frac{1}{3} M v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

.

Connects microscopic speed distribution parameter with thermodynamic temperature and molar mass.

Key Concept

Kinetic Theory of Matter and Pressure of Gases

Question 116Question →

In an experiment to determine the specific latent heat of vaporization of water at $100^\circ\text{C}$ using an electric immersion heater in an uninsulated vessel, if heat loss from the vessel to the cooler surrounding environment is neglected in the calculations, the experimentally determined value of the specific latent heat of vaporization will be higher than the true value.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is correct because the measured energy supplied by the heater includes both the energy required for the phase change and the thermal energy dissipated to the surroundings. Dividing this larger total energy by the mass of vaporized liquid results in an overestimation of the specific latent heat of vaporization.

Step-by-Step Solution

1

Formulate the thermal energy balance equation including heat loss.

Q_{\text{electrical}} = Q_{\text{vaporization}} + Q_{\text{loss}}

, where

Q_{\text{electrical}} = P \cdot t

and

Q_{\text{vaporization}} = m \cdot L_{\text{true}}

.

Energy conservation dictates that the electrical energy supplied by the heater equals the thermal energy used for vaporization plus the heat lost to the cooler ambient surroundings.

2

Express the calculated specific latent heat

L_{\text{calc}}

under the assumption of zero heat loss.

L_{\text{calc}} = \frac{Q_{\text{electrical}}}{m} = \frac{m \cdot L_{\text{true}} + Q_{\text{loss}}}{m}

.

The experimenter measures the total electrical power and time, assuming all of this energy converted liquid into vapor.

3

Compare the calculated specific latent heat

L_{\text{calc}}

with the true value

L_{\text{true}}

.

L_{\text{calc}} = L_{\text{true}} + \frac{Q_{\text{loss}}}{m} > L_{\text{true}}

.

Because

Q_{\text{loss}} > 0

and mass

m > 0

, the ratio

\frac{Q_{\text{loss}}}{m}

is positive, meaning

L_{\text{calc}}

is higher than

L_{\text{true}}

.

Key Concept

Experimental Heat Loss Error Propagation in Latent Heat Calculations

Question 117Question →

A solid metallic sphere of mass $0.4\text{ kg}$ and specific heat capacity $500\text{ J kg}^{-1}\text{ K}^{-1}$ is heated to $100^\circ\text{C}$ and then placed into a well-insulated calorimeter of heat capacity $100\text{ J K}^{-1}$ . The calorimeter contains $0.5\text{ kg}$ of a liquid initially at $20^\circ\text{C}$ . If the final equilibrium temperature of the system is $40^\circ\text{C}$ and heat loss to the surroundings is negligible, what is the specific heat capacity of the liquid?

1200\text{ J kg}^{-1}\text{ K}^{-1}

1000\text{ J kg}^{-1}\text{ K}^{-1}

1100\text{ J kg}^{-1}\text{ K}^{-1}

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

Show answer & explanation

Answer: $1000\text{ J kg}^{-1}\text{ K}^{-1}$

Answer

1000\text{ J kg}^{-1}\text{ K}^{-1}

The heat lost by the cooling metallic sphere is

Q = 0.4 \times 500 \times 60 = 12000\text{ J}

. This energy raises the temperature of both the calorimeter container and the liquid by

20^\circ\text{C}

. Setting

(100 + 0.5 c_{\text{liquid}}) \times 20 = 12000

yields

100 + 0.5 c_{\text{liquid}} = 600

, giving

c_{\text{liquid}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the total heat energy lost by the cooling metallic sphere.

Q_{\text{lost}} = m_1 c_1 (T_{\text{initial}} - T_{\text{final}}) = 0.4 \times 500 \times (100 - 40) = 12000\text{ J}

Heat lost depends on mass, specific heat capacity, and temperature decrease of the hot body.

2

Formulate the thermal energy absorption by the calorimeter and liquid.

Q_{\text{gained}} = (C_{\text{cal}} + m_2 c_2) (T_{\text{final}} - T_{\text{initial, liquid}}) = (100 + 0.5 c_2) \times (40 - 20)

Heat capacity of the container (

C_{\text{cal}}

) is an extensive property already accounting for container mass, whereas specific heat capacity (

c_2

) must be multiplied by liquid mass.

3

Equate heat lost to heat gained using conservation of thermal energy and solve for

c_2

.

12000 = (100 + 0.5 c_2) \times 20 \implies 600 = 100 + 0.5 c_2 \implies 0.5 c_2 = 500 \implies c_2 = 1000\text{ J kg}^{-1}\text{ K}^{-1}

Assuming no external losses, energy conservation dictates that thermal energy lost equals thermal energy absorbed.

Key Concept

Method of Mixtures and Distinction Between Heat Capacity and Specific Heat Capacity

Question 118Question →

A closed rigid container holds a mixture of dry air and water vapour at a temperature of $27^\circ\text{C}$ under a total pressure of $740\text{ mmHg}$ . The relative humidity of the air inside the container is $80\%$ , and the saturated vapour pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If the container is heated at constant volume to $127^\circ\text{C}$ , what is the partial pressure of the dry air in $\text{mmHg}$ at this higher temperature?

Show answer & explanation

Answer: 960

Answer

The partial pressure of dry air inside the container at

127^\circ\text{C}

is

960\text{ mmHg}

.

First, the partial pressure of water vapour at

27^\circ\text{C}

is determined by multiplying relative humidity (

80\%

) by the saturated vapour pressure (

25\text{ mmHg}

), yielding

20\text{ mmHg}

. Next, subtracting this vapour pressure from the total pressure of

740\text{ mmHg}

gives the partial pressure of dry air alone as

720\text{ mmHg}

at

27^\circ\text{C}

(

300\text{ K}

). Finally, applying the Pressure Law (

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

) for the dry air between

300\text{ K}

and

400\text{ K}

(

127^\circ\text{C}

) yields

P_2 = 720 \times \frac{400}{300} = 960\text{ mmHg}

.

Step-by-Step Solution

1

Calculate the partial pressure of water vapour at

27^\circ\text{C}

P_{\text{vapour}, 1} = 0.80 \times 25\text{ mmHg} = 20\text{ mmHg}

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the initial partial pressure of the dry air at

27^\circ\text{C}

using Dalton's Law

P_{\text{dry air}, 1} = 740\text{ mmHg} - 20\text{ mmHg} = 720\text{ mmHg}

Total pressure of a gas mixture is the sum of the partial pressures of its individual components.

3

Convert temperatures to Kelvin and apply Gay-Lussac's Pressure Law for dry air at constant volume

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

,

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

.

P_{\text{dry air}, 2} = 720\text{ mmHg} \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 960\text{ mmHg}

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

Key Concept

Dalton's Law of Partial Pressures and Gay-Lussac's Pressure Law applied to gas-vapour mixtures

Question 119Question →

A metal block of mass $4.0\text{ kg}$ absorbs $12\text{ kJ}$ of thermal energy, causing its temperature to rise by $15\text{ K}$ . What is the specific heat capacity of the metal block in $\text{J kg}^{-1}\text{K}^{-1}$ ?

Show answer & explanation

Answer: 200

Answer

The specific heat capacity of the metal block is

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

.

Specific heat capacity

c

is determined using

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

. Substituting

Q = 12,000\text{ J}

,

m = 4.0\text{ kg}

, and

\Delta T = 15\text{ K}

yields

c = \frac{12000}{4.0 \times 15} = 200\text{ J kg}^{-1}\text{K}^{-1}

.

Step-by-Step Solution

1

Convert given energy to standard SI units

Q = 12\text{ kJ} = 12,000\text{ J}

Calculation of specific heat capacity requires heat energy in Joules.

2

Relate heat energy, mass, specific heat capacity, and temperature change

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

Specific heat capacity

c

represents heat energy per unit mass per Kelvin temperature change.

3

Substitute the values and compute the result

c = \frac{12000}{4.0 \times 15} = 200\text{ J kg}^{-1}\text{K}^{-1}

Evaluating the expression gives

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

.

Key Concept

Specific heat capacity is the quantity of heat required to raise the temperature of

1\text{ kg}

of a substance by

1\text{ K}

, expressed as

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

.

Question 120Question →

A sealed room with a volume of $50\text{ m}^3$ at a temperature of $20^\circ\text{C}$ contains $0.40\text{ kg}$ of water vapour. If the mass of water vapour required to saturate $1\text{ m}^3$ of air at $20^\circ\text{C}$ is $0.016\text{ kg}$ , what is the relative humidity of the air in the room?

25\%

40\%

50\%

80\%

Show answer & explanation

Answer: $50\%$

Answer

50\%

Relative humidity is defined as the ratio of the actual mass of water vapour present in a given volume of air to the mass of water vapour required to saturate the same volume at the same temperature. For

50\text{ m}^3

of air, the mass needed for saturation is

50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

. Dividing the actual mass (

0.40\text{ kg}

) by

0.80\text{ kg}

and multiplying by

100\%

gives

50\%

.

Step-by-Step Solution

1

Calculate the total mass of water vapour required to saturate the entire room volume.

\text{Total saturation mass} = 50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

The saturation density gives the maximum water vapour

1\text{ m}^3

can hold, so it must be scaled by the room volume.

2

Apply the formula for relative humidity.

\text{Relative Humidity} = \left(\frac{\text{Actual mass of water vapour}}{\text{Saturation mass of water vapour}}\right) \times 100\%

Relative humidity measures the degree of saturation of an air sample at a specific temperature.

3

Substitute the known values to find the relative humidity.

\text{Relative Humidity} = \left(\frac{0.40\text{ kg}}{0.80\text{ kg}}\right) \times 100\% = 50\%

Simplifying the fraction

\frac{0.40}{0.80} = 0.5

, which equals

50\%

.

Key Concept

Relative Humidity

Estimated Time:1m 15s

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