Modes of Heat Transfer (Conduction, Convection, and Radiation) Practice Questions — JAMB UTME

Question 1Question →

Match each vacuum flask component or surface feature on the left with its primary mechanism for controlling heat transfer on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Silvered inner surfaces match with radiation reflection; evacuated space matches with elimination of conduction and convection; cork stopper matches with prevention of convection and conduction at the opening; dull black casing matches with maximizing thermal radiation emission/absorption.

Each feature of the vacuum flask targets a specific heat transfer mode: silvering reflects infrared radiation; the vacuum eliminates particle-dependent transfer (conduction and convection); cork acts as an insulator preventing convection and conduction at the top; and black surfaces maximize thermal radiation emission/absorption.

Step-by-Step Solution

1

Analyze the silvered inner walls of a vacuum flask

Silver surfaces are good reflectors of heat rays (infrared waves).

Radiant heat travels via electromagnetic waves and is reflected by shiny metallic coatings, minimizing radiation heat loss.

2

Analyze the vacuum space between the glass walls

Conduction and convection cannot occur across a vacuum.

Both conduction (particle vibration/electron flow) and convection (fluid movement) strictly require a physical medium.

3

Analyze the cork/plastic stopper

Cork prevents hot air circulation and thermal conduction across the opening.

Cork is a poor conductor of heat and stops evaporative/convective air currents from leaving the container.

4

Analyze the dull black exterior

Dull black surfaces are efficient radiation emitters/absorbers.

According to radiation principles, black matte surfaces radiate energy much faster than polished surfaces.

Key Concept

Modes of Heat Transfer and Practical Applications in Thermal Insulation

Question 2Question →

A uniform metal rod of length $0.50\text{ m}$ and cross-sectional area $2.0 \times 10^{-3}\text{ m}^2$ has a thermal conductivity of $400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . If a temperature difference of $50\text{ K}$ is maintained between its ends, what is the rate of heat flow through the rod?

80\text{ W}

40\text{ W}

20\text{ W}

160\text{ W}

Show answer & explanation

Answer: $80\text{ W}$

Answer

The rate of heat flow through the rod is

80\text{ W}

.

The rate of heat transfer by conduction is calculated using Fourier's law,

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

. Substituting

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 2.0 \times 10^{-3}\text{ m}^2

,

\Delta T = 50\text{ K}

, and

d = 0.50\text{ m}

yields

80\text{ W}

.

Step-by-Step Solution

1

Identify the given parameters for thermal conduction.

Thermal conductivity

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, cross-sectional area

A = 2.0 \times 10^{-3}\text{ m}^2

, temperature difference

\Delta T = 50\text{ K}

, and length

d = 0.50\text{ m}

.

These parameters form the components of Fourier's law of thermal conduction.

2

Apply the conduction rate formula

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

.

\frac{Q}{t} = \frac{400 \times (2.0 \times 10^{-3}) \times 50}{0.50}

Heat transfer per unit time depends directly on conductivity, area, and temperature difference, and inversely on thickness/length.

3

Calculate the numerical value.

\frac{Q}{t} = \frac{40}{0.50} = 80\text{ W}

Dividing the product of the numerator terms (

40\text{ J/s}\cdot\text{m}

) by length (

0.50\text{ m}

) yields the rate of energy transfer.

Key Concept

Thermal Conduction Rate (Fourier's Law)

Estimated Time:45s

Question 3Question →

A composite furnace wall consists of two tightly joined layers of equal cross-sectional area. Layer 1 has a thickness of $0.02\text{ m}$ and thermal conductivity of $200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . Layer 2 has a thickness of $0.04\text{ m}$ and thermal conductivity of $100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . The outer surface of Layer 1 is maintained at $120^\circ\text{C}$ and the outer surface of Layer 2 is maintained at $20^\circ\text{C}$ . Under steady-state conditions, what is the rate of heat transfer per unit area through the composite wall?

200\text{ kW}\cdot\text{m}^{-2}

250\text{ kW}\cdot\text{m}^{-2}

500\text{ kW}\cdot\text{m}^{-2}

100\text{ kW}\cdot\text{m}^{-2}

Show answer & explanation

Answer: $200\text{ kW}\cdot\text{m}^{-2}$

Answer

The rate of heat transfer per unit area through the composite furnace wall is

200\text{ kW}\cdot\text{m}^{-2}

.

Under steady-state conduction through composite layers in series, the total thermal resistance per unit area is the sum of the individual thermal resistances:

R_{\text{total}} = \frac{d_1}{k_1} + \frac{d_2}{k_2} = 1.0 \times 10^{-4} + 4.0 \times 10^{-4} = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

. Applying Fourier's law of heat conduction, the heat flux is

\frac{\Delta T}{R_{\text{total}}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}

.

Step-by-Step Solution

1

Calculate the thermal resistance per unit area (

R

) for each layer using

R_i = \frac{d_i}{k_i}

.

R_1 = \frac{0.02\text{ m}}{200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 1.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

and

R_2 = \frac{0.04\text{ m}}{100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 4.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

.

Thermal resistance quantifies a material's opposition to conductive heat flow per unit cross-sectional area.

2

Find the total thermal resistance per unit area (

R_{\text{total}}

) by summing the series resistances.

R_{\text{total}} = R_1 + R_2 = (1.0 \times 10^{-4}) + (4.0 \times 10^{-4}) = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

.

Layers in series add their thermal resistances directly.

3

Determine the rate of heat flow per unit area

\frac{Q/t}{A} = \frac{\Delta T}{R_{\text{total}}}

.

\frac{Q/t}{A} = \frac{120^\circ\text{C} - 20^\circ\text{C}}{5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}

.

At steady state, the heat flux across the composite structure is driven by the overall temperature gradient divided by total thermal resistance.

Key Concept

Thermal Resistance in Composite Layers (Conduction)

Estimated Time:2m 0s

Question 4Question →

A glass window pane has an area of $1.50\text{ m}^2$ and a thickness of $4.0\text{ mm}$ . The inner and outer surface temperatures of the glass are maintained at $25^\circ\text{C}$ and $5^\circ\text{C}$ respectively. If the thermal conductivity of glass is $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , what is the rate of heat transfer through the glass pane by conduction?

6000\text{ W}

4000\text{ W}

87900\text{ W}

6.0\text{ W}

Show answer & explanation

Answer: $6000\text{ W}$

Answer

The rate of heat transfer through the glass pane is

6000\text{ W}

.

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

\frac{Q}{t}

is given by

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

. Substituting

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 1.50\text{ m}^2

,

T_1 - T_2 = 20\text{ K}

, and

d = 0.004\text{ m}

gives

\frac{Q}{t} = \frac{0.80 \times 1.50 \times 20}{0.004} = 6000\text{ W}

.

Step-by-Step Solution

1

Identify the given values and convert all quantities to SI units.

Area

A = 1.50\text{ m}^2

, thermal conductivity

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, thickness

d = 4.0\text{ mm} = 4.0 \times 10^{-3}\text{ m}

, and temperature difference

\Delta T = 25^\circ\text{C} - 5^\circ\text{C} = 20\text{ K}

.

Fourier's law requires all linear dimensions to be in meters and temperatures in kelvins/degrees Celsius consistently.

2

Apply Fourier's law of thermal conduction equation.

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

The rate of heat conduction is directly proportional to thermal conductivity, surface area, and temperature difference, and inversely proportional to material thickness.

3

Substitute the values into the formula and solve.

\frac{Q}{t} = \frac{0.80 \times 1.50 \times 20}{4.0 \times 10^{-3}} = \frac{24}{0.004} = 6000\text{ W}

Calculates the total heat energy conducted per second across the window pane.

Key Concept

Thermal Conduction Rate Equation

Estimated Time:1m 30s

Question 5Question →

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 6Question →

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 7Question →

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 8Question →

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 9Question →

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 10Question →

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 11Question →

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 12Question →

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 13Question →

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 14Question →

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 15Question →

Match each physical component or thermal phenomenon listed on the left with its corresponding primary heat transfer mechanism and operational principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: the evacuated space matches the prevention of conduction and convection due to lack of medium; silvered inner glass matches the minimization of radiation via reflective low-emissivity surfaces; the thick copper base matches enhanced conduction via free electron diffusion; and the offshore land breeze matches natural convection driven by air density differences.

Each phenomenon is matched correctly to its fundamental physical requirement: vacuum interspace eliminates conduction and convection by removing matter; silvered surfaces prevent radiation loss via reflection; copper base accelerates conduction through electron movement; land breeze is a fluid density-driven convection current.

Step-by-Step Solution

1

Analyze the vacuum interspace mechanism

Since conduction requires direct particle collisions and convection requires fluid flow, removing air creates a vacuum that eliminates both conduction and convection.

Both conduction and convection depend on a physical medium.

2

Analyze the silvered glass walls function

Radiation does not require a medium and is governed by surface emissivity. Shiny, silvered surfaces reflect thermal radiation back into the vessel.

Polished metallic coatings decrease thermal radiation emissivity and increase reflectivity.

3

Analyze the copper cooking base conduction property

Metals like copper transfer heat rapidly across solid structures using free electrons alongside atomic lattice vibrations.

Free electron diffusion makes copper an exceptionally good conductor of heat.

4

Analyze the coastal land breeze phenomenon

At night, land loses thermal energy faster than water, causing cooler dense air above land to slide under warmer rising air over the sea.

This fluid circulation driven by thermal expansion and buoyancy differences is natural convection.

Key Concept

Distinct physical requirements and microscopic mechanisms of conduction, convection, and thermal radiation.

Estimated Time:2m 0s

Question 16Question →

A glass window pane of thickness $4.0\text{ mm}$ and surface area $1.5\text{ m}^2$ maintains an inner surface temperature of $20^\circ\text{C}$ and an outer surface temperature of $5^\circ\text{C}$ . If the thermal conductivity of glass is $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , what is the rate of heat transfer by conduction through the window?

4500\text{ W}

4.5\text{ W}

86,400\text{ W}

0.072\text{ W}

Show answer & explanation

Answer: $4500\text{ W}$

Answer

4500\text{ W}

(or

4.5\text{ kW}

)

The rate of conductive heat transfer is governed by Fourier's law of thermal conduction:

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

. Substituting

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 1.5\text{ m}^2

, temperature difference

\Delta T = 15\text{ K}

, and thickness

d = 0.004\text{ m}

gives

\frac{Q}{t} = \frac{0.80 \times 1.5 \times 15}{0.004} = 4500\text{ W}

.

Step-by-Step Solution

1

Identify the given parameters and convert units to standard SI units

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 1.5\text{ m}^2

,

\Delta T = 20^\circ\text{C} - 5^\circ\text{C} = 15\text{ K}

, and

d = 4.0\text{ mm} = 4.0 \times 10^{-3}\text{ m}

.

Thermal conductivity formulas require distance/thickness in meters.

2

Apply the law of thermal conduction formula

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

The rate of heat transfer by conduction is directly proportional to thermal conductivity, surface area, and temperature difference, and inversely proportional to thickness.

3

Substitute the values and calculate the heat transfer rate

\frac{Q}{t} = \frac{0.80 \times 1.5 \times 15}{4.0 \times 10^{-3}} = \frac{18}{0.004} = 4500\text{ W}

Performing accurate arithmetic yields

4500\text{ Joules per second}

(Watts).

Key Concept

Rate of thermal conduction through a uniform slab

Question 17Question →

A double-glazed window of total surface area $1.5\text{ m}^2$ consists of two glass panes, each of thickness $4.0\text{ mm}$ and thermal conductivity $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , separated by a stagnant air gap of thickness $2.0\text{ mm}$ with thermal conductivity $0.025\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . If a steady-state temperature difference of $18.0^\circ\text{C}$ is maintained across the window's outer boundary surfaces, what is the rate of heat transfer through the window in watts?

Show answer & explanation

Answer: 300

Answer

The steady-state rate of heat transfer through the double-glazed window is

300\text{ W}

.

Heat conduction through composite layers in series is governed by the total thermal resistance. The thermal resistance per unit area of each layer is

r = \frac{d}{k}

. For two

4.0\text{ mm}

glass panes (

r = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

each) and one

2.0\text{ mm}

air gap (

r = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

), the total unit resistance is

r_{\text{total}} = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

. Multiplying by the area

1.5\text{ m}^2

and temperature difference

18.0^\circ\text{C}

gives

\frac{Q}{t} = \frac{1.5 \times 18.0}{0.090} = 300\text{ W}

.

Step-by-Step Solution

1

Convert layer thicknesses to standard units (meters)

d_{\text{glass}} = 4.0\text{ mm} = 0.004\text{ m}

,

d_{\text{air}} = 2.0\text{ mm} = 0.002\text{ m}

SI units are required for calculations using thermal conductivity in

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

.

2

Calculate thermal resistance per unit area for each layer

r_{\text{glass}} = \frac{0.004}{0.80} = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

,

r_{\text{air}} = \frac{0.002}{0.025} = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

Thermal resistance per unit area is given by

r = \frac{d}{k}

.

3

Sum thermal resistances in series to obtain total unit resistance

r_{\text{total}} = r_{\text{glass1}} + r_{\text{air}} + r_{\text{glass2}} = 0.005 + 0.080 + 0.005 = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

Heat flows sequentially through all three layers in series.

4

Calculate rate of heat flow across total window area

\frac{Q}{t} = \frac{A \cdot \Delta T}{r_{\text{total}}} = \frac{1.5 \cdot 18.0}{0.090} = 300\text{ W}

Rate of thermal conduction through composite series layers is

\frac{Q}{t} = \frac{\Delta T}{R_{\text{total}}}

where

R_{\text{total}} = \frac{r_{\text{total}}}{A}

.

Key Concept

Series thermal conduction through composite layers and thermal resistance

Question 18Question →

A metal boiler base has a thermal conductivity of $200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , a thickness of $6.0\text{ mm}$ , and an effective surface area of $0.15\text{ m}^2$ . If heat is conducted through the base at a rate of $500\text{ kW}$ , what is the temperature difference across the two faces of the boiler base?

10\text{ }^\circ\text{C}

100\text{ }^\circ\text{C}

373\text{ }^\circ\text{C}

2.5\text{ }^\circ\text{C}

Show answer & explanation

Answer: $100\text{ }^\circ\text{C}$

Answer

The temperature difference across the two faces of the boiler base is

100\text{ }^\circ\text{C}

.

Using Fourier's law of thermal conduction,

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

. Substituting

P = 500,000\text{ W}

,

d = 0.006\text{ m}

,

k = 200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, and

A = 0.15\text{ m}^2

yields

\Delta T = \frac{500,000 \times 0.006}{200 \times 0.15} = 100\text{ }^\circ\text{C}

.

Step-by-Step Solution

1

Convert given quantities into standard SI units

Heat transfer rate

P = 500\text{ kW} = 500,000\text{ W}

, thickness

d = 6.0\text{ mm} = 0.006\text{ m}

, area

A = 0.15\text{ m}^2

, thermal conductivity

k = 200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

All parameters must be in SI base units before substitution into the thermal conduction equation.

2

State the equation for rate of heat conduction and rearrange for temperature difference (

\Delta T

)

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

.

To isolate the unknown temperature gradient component.

3

Substitute the values and compute the result

\Delta T = \frac{500,000 \times 0.006}{200 \times 0.15} = \frac{3000}{30} = 100\text{ }^\circ\text{C}.

Direct algebraic simplification yields the temperature difference.

Key Concept

Thermal Conduction Rate Formula

Estimated Time:1m 30s

Question 19Question →

A uniform metal cylinder of thermal conductivity $400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ has a cross-sectional area of $2.0 \times 10^{-3}\text{ m}^2$ and a length of $0.50\text{ m}$ . Heat flows axially through the cylinder at a steady rate of $160\text{ W}$ . Assuming no thermal losses through the lateral surface, what is the temperature difference between the two ends of the cylinder?

100\text{ K}

200\text{ K}

400\text{ K}

50\text{ K}

Show answer & explanation

Answer: $100\text{ K}$

Answer

The temperature difference between the two ends of the cylinder is

100\text{ K}

.

According to Fourier's law of heat conduction, the rate of heat transfer through a material is given by

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

. Rearranging the equation to solve for the temperature difference gives

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

. Substituting

Q/t = 160\text{ W}

,

d = 0.50\text{ m}

,

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, and

A = 2.0 \times 10^{-3}\text{ m}^2

yields

\Delta T = \frac{160 \times 0.50}{400 \times 2.0 \times 10^{-3}} = \frac{80}{0.8} = 100\text{ K}

.

Step-by-Step Solution

1

Identify the given thermal parameters and the rate of heat flow equation.

Rate of heat flow

\frac{Q}{t} = 160\text{ W}

, thermal conductivity

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, cross-sectional area

A = 2.0 \times 10^{-3}\text{ m}^2

, length

d = 0.50\text{ m}

. Formula:

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

.

Conduction through a solid body under steady-state conditions obeys Fourier's law of thermal conduction.

2

Rearrange the conduction formula to solve for the temperature difference

\Delta T

.

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

.

Isolating

\Delta T

requires multiplying both sides by length

d

and dividing by

k A

.

3

Substitute the values and compute

\Delta T

.

\Delta T = \frac{160 \times 0.50}{400 \times 2.0 \times 10^{-3}} = \frac{80}{0.8} = 100\text{ K}

.

Performing accurate arithmetic yields the temperature difference.

Key Concept

Thermal Conduction Rate

Question 20Question →

A spherical black body radiator with an initial radius of $0.10\text{ m}$ emits thermal radiation at a rate of $E_1$ when its surface temperature is $400\text{ K}$ . If the radius of the sphere is doubled to $0.20\text{ m}$ and its absolute temperature is reduced to $200\text{ K}$ , what is the value of the ratio of the new rate of heat radiation to the initial rate, $\frac{E_2}{E_1}$ ?

Show answer & explanation

Answer: 0.25

Answer

The ratio of the new rate of heat radiation to the initial rate is

0.25

.

According to the Stefan-Boltzmann law, the rate of thermal radiation emitted by a black body is directly proportional to its surface area (

A \propto r^2

) and the fourth power of its absolute temperature (

T^4

). Doubling the radius increases the surface area by a factor of

2^2 = 4

, while halving the absolute temperature decreases the radiation rate per unit area by a factor of

(1/2)^4 = 1/16

. The net ratio of the new emission rate to the initial emission rate is

4 \times (1/16) = 0.25

.

Step-by-Step Solution

1

Apply Stefan-Boltzmann's law of radiation to formulate the rate of heat emission.

The total power radiated by a sphere of radius

r

at thermodynamic temperature

T

is given by

P = \sigma A T^4 = 4\pi \sigma r^2 T^4

, where

\sigma

is the Stefan-Boltzmann constant.

Thermal radiation emission rate depends directly on surface area (

A = 4\pi r^2

) and the fourth power of absolute temperature (

T^4

).

2

Formulate the ratio

\frac{E_2}{E_1}

using the scaled physical parameters.

\frac{E_2}{E_1} = \frac{4\pi \sigma r_2^2 T_2^4}{4\pi \sigma r_1^2 T_1^4} = \left(\frac{r_2}{r_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4

Constants

4\pi\sigma

cancel out when evaluating relative changes.

3

Substitute the given numerical ratios into the equation and compute the result.

Given

\frac{r_2}{r_1} = \frac{0.20}{0.10} = 2

and

\frac{T_2}{T_1} = \frac{200}{400} = 0.5

, we get

\frac{E_2}{E_1} = (2)^2 \times (0.5)^4 = 4 \times \frac{1}{16} = 0.25

.

Evaluating

2^2 = 4

and

(0.5)^4 = 1/16

yields a final ratio of

4/16 = 0.25

.

Key Concept

Stefan-Boltzmann Law of Radiation

Modes of Heat Transfer (Conduction, Convection, and Radiation)

Answer

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