Thermal Physics Practice Questions — JAMB UTME — Page 7

Question 121Question →

If a fixed mass of unsaturated air in a closed container is cooled at constant total pressure, its relative humidity increases until it reaches the dew point; upon cooling further below the dew point, condensation occurs such that the relative humidity remains at 100% while the saturated vapour pressure continues to decrease.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Cooling unsaturated air reduces its maximum moisture-holding capacity (SVP). Before saturation, actual vapour pressure is fixed, so relative humidity rises to 100% at the dew point. Continuous cooling past this point forces condensation, decreasing the actual vapour pressure alongside the SVP to keep the air continuously saturated at 100% relative humidity.

Step-by-Step Solution

1

Analyze the relationship between cooling and relative humidity for unsaturated air.

As temperature decreases, the saturated vapour pressure (SVP) decreases while the actual partial vapour pressure remains constant, causing relative humidity (actual VP / SVP × 100%) to increase.

Relative humidity is inversely proportional to the saturated vapour pressure at a fixed moisture content.

2

Identify the state reached when relative humidity reaches 100%.

The air reaches saturation at the dew point temperature where actual vapour pressure equals SVP.

By definition, the dew point is the temperature at which water vapour in air begins to condense.

3

Determine the thermodynamic behavior during continuous cooling below the dew point.

Excess vapour condenses into liquid water, decreasing actual vapour pressure to continuously match the lower SVP at each reduced temperature.

Air cannot maintain an unsaturated or supersaturated equilibrium state in the presence of condensate, keeping relative humidity locked at 100%.

Key Concept

Vapour saturation, dew point determination, and condensation mechanics during air cooling.

Estimated Time:1m 30s

Question 122Question →

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 123Question →

If two solid spheres, $A$ and $B$ , constructed from the same uniform metallic material, have radii in the ratio $2:1$ respectively, then supplying equal quantities of heat energy to both spheres will cause sphere $B$ to experience a temperature rise eight times that of sphere $A$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the mass and heat capacity of a uniform solid sphere scale with the cube of its radius (

r^3

), giving sphere

A

eight times the heat capacity of sphere

B

. For an equal input of thermal energy, sphere

B

undergoes eight times the temperature rise of sphere

A

.

The statement is correct because volume scales as

r^3

, giving sphere

A

eight times the mass and heat capacity of sphere

B

. As a result, sphere

B

undergoes eight times the temperature increase of sphere

A

when absorbing equal heat energy.

Step-by-Step Solution

1

Relate the masses of the two spheres using their radii ratio.

m_A = \rho V_A = \rho \cdot \frac{4}{3}\pi (2r_B)^3 = 8 \left(\rho \cdot \frac{4}{3}\pi r_B^3\right) = 8 m_B

.

Mass is proportional to volume for uniform density, and volume scales as

r^3

.

2

Express the heat capacities of both spheres.

C_A = m_A c = 8 m_B c = 8 C_B

.

Heat capacity

C = mc

is an extensive property proportional to mass, while specific heat capacity

c

is constant for a given material.

3

Compare the temperature rises for equal heat energy

Q

.

\Delta T_B = \frac{Q}{C_B} = \frac{Q}{\frac{1}{8}C_A} = 8 \left(\frac{Q}{C_A}\right) = 8 \Delta T_A

.

Temperature rise

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

is inversely proportional to heat capacity when heat input

Q

is identical.

Key Concept

Extensive nature of heat capacity and its scaling with volume (

r^3

) versus intensive specific heat capacity.

Estimated Time:1m 30s

Question 124Question →

Calculate the thermal energy, in joules, required to completely melt $0.50\text{ kg}$ of ice at its melting point of $0^\circ\text{C}$ . (Take the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J kg}^{-1}$ ).

Show answer & explanation

Answer: 168000

Answer

The total heat required to completely melt the ice is

168,000\text{ J}

.

During a state change from solid to liquid at constant temperature, the thermal energy absorbed is given directly by

Q = m L_f

. Multiplying mass (

0.50\text{ kg}

) by specific latent heat (

3.36 \times 10^5\text{ J kg}^{-1}

) gives

168,000\text{ J}

.

Step-by-Step Solution

1

Identify the given values and formula

Mass

m = 0.50\text{ kg}

, Specific latent heat of fusion

L_f = 3.36 \times 10^5\text{ J kg}^{-1}

, Formula:

Q = m L_f

Since the phase change occurs at constant temperature (

0^\circ\text{C}

), sensible heat calculation (

mc\Delta T

) is zero and only latent heat is needed.

2

Substitute values into the formula and calculate

Q = 0.50\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 168,000\text{ J}

Direct multiplication of mass and specific latent heat yields the energy required in Joules.

Key Concept

Specific Latent Heat of Fusion

Question 125Question →

An electric heater rated at $500\text{ W}$ is used to melt $0.20\text{ kg}$ of ice initially at $0^\circ\text{C}$ and heat the resulting water to $20^\circ\text{C}$ . Assuming no heat losses to the surroundings, what is the time required in seconds? (Specific latent heat of fusion of ice = $3.36 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water = $4.2 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}$ )

Show answer & explanation

Answer: 168

Answer

168 s

The total energy required is the sum of the heat required for melting (

mL_f = 67,200\text{ J}

) and the heat required for warming the liquid (

mc\Delta T = 16,800\text{ J}

), giving

84,000\text{ J}

. Dividing this by the heater power rating of

500\text{ W}

gives

168\text{ s}

.

Step-by-Step Solution

1

Calculate thermal energy needed to change phase from ice to water at

0^\circ\text{C}

Q_1 = m L_f = 0.20\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 67,200\text{ J}

Latent heat of fusion accounts for phase change at constant temperature.

2

Calculate thermal energy needed to raise water temperature from

0^\circ\text{C}

to

20^\circ\text{C}

Q_2 = m c \Delta T = 0.20\text{ kg} \times 4.2 \times 10^3\text{ J kg}^{-1}\text{K}^{-1} \times 20\text{ K} = 16,800\text{ J}

Sensible heat increases kinetic energy of liquid water molecules.

3

Sum energy required for both phase change and temperature change

Q_{\text{total}} = Q_1 + Q_2 = 67,200\text{ J} + 16,800\text{ J} = 84,000\text{ J}

Total energy is the sum of sensible heat and latent heat.

4

Calculate heating time using heater power rating

t = \frac{Q_{\text{total}}}{P} = \frac{84,000\text{ J}}{500\text{ W}} = 168\text{ s}

Power is defined as energy transferred per unit time (

P = \frac{Q}{t}

).

Key Concept

Energy balance combining latent heat and sensible heat

Estimated Time:2m 0s

Question 126Question →

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 127Question →

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 128Question →

An electric heater rated at $1.0\text{ kW}$ is immersed in $0.20\text{ kg}$ of water at its boiling point of $100^\circ\text{C}$ . Assuming all the energy supplied by the heater is used to vaporize the water and there is no heat loss to the surroundings, how long will it take for all the water to completely turn into steam? (Take the specific latent heat of vaporization of water $L_v = 2.25 \times 10^6\text{ J kg}^{-1}$ )

450\text{ s}

45,000\text{ s}

84\text{ s}

2250\text{ s}

Show answer & explanation

Answer: $450\text{ s}$

Answer

The total time required to completely vaporize the water is

450\text{ s}

.

At the boiling point of

100^\circ\text{C}

, all thermal energy supplied goes into changing the liquid water to steam without increasing its temperature. The required heat is

Q = m L_v = 0.20\text{ kg} \times (2.25 \times 10^6\text{ J kg}^{-1}) = 450,000\text{ J}

. Dividing energy by the heater power (

1000\text{ W}

) yields

450\text{ s}

.

Step-by-Step Solution

1

Calculate the total thermal energy required to vaporize the water at constant temperature.

Q = m L_v = 0.20\text{ kg} \times (2.25 \times 10^6\text{ J kg}^{-1}) = 4.5 \times 10^5\text{ J} = 450,000\text{ J}

Phase change at the boiling point occurs at constant temperature, so only latent heat is required.

2

Convert the power of the heater from kilowatts to watts.

P = 1.0\text{ kW} = 1000\text{ W} = 1000\text{ J s}^{-1}

Standard SI unit for electrical power is watts (joules per second).

3

Determine the time required using the relationship between energy, power, and time.

t = \frac{Q}{P} = \frac{450,000\text{ J}}{1000\text{ W}} = 450\text{ s}

Power is the rate of energy transfer (

P = \frac{Q}{t}

).

Key Concept

Latent heat of vaporization and electrical power heat transfer at constant temperature during a phase change.

Estimated Time:1m 30s

Question 129Question →

The table below presents the mass, quantity of heat absorbed, and resulting temperature change for three solid metal blocks, $P$ , $Q$ , and $R$ :

Block	Mass ( $\text{kg}$ )	Heat Absorbed ( $\text{J}$ )	Temperature Change ( $\text{K}$ )
$P$	$0.20$	$800$	$10$
$Q$	$0.50$	$1500$	$15$
$R$	$0.40$	$1200$	$20$

Which of the following statements correctly compares the thermal properties of the blocks?

Block

P

has the highest specific heat capacity, while block

Q

has the highest total heat capacity.

Block

Q

has both the highest specific heat capacity and the highest total heat capacity.

Block

R

has the highest specific heat capacity, while block

P

has the lowest total heat capacity.

Block

P

has the lowest specific heat capacity, while block

R

has the highest total heat capacity.

Show answer & explanation

Answer: Block $P$ has the highest specific heat capacity, while block $Q$ has the highest total heat capacity.

Answer

Block

P

has the highest specific heat capacity, while block

Q

has the highest total heat capacity.

The total heat capacity

C

is found using

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

, giving

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

for

P

,

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

for

Q

, and

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

for

R

. Dividing heat capacity by mass yields the specific heat capacity

c = \frac{C}{m}

, resulting in

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

for

P

,

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

for

Q

, and

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

for

R

. Therefore, block

P

has the highest specific heat capacity and block

Q

has the highest total heat capacity.

Step-by-Step Solution

1

Calculate the total heat capacity (

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

) for each block

C_P = \frac{800}{10} = 80\text{ J K}^{-1}

,

C_Q = \frac{1500}{15} = 100\text{ J K}^{-1}

,

C_R = \frac{1200}{20} = 60\text{ J K}^{-1}

Heat capacity measures the total thermal energy required to raise the entire object's temperature by

1\text{ K}

.

2

Calculate the specific heat capacity (

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

) for each block

c_P = \frac{80}{0.20} = 400\text{ J kg}^{-1}\text{ K}^{-1}

,

c_Q = \frac{100}{0.50} = 200\text{ J kg}^{-1}\text{ K}^{-1}

,

c_R = \frac{60}{0.40} = 150\text{ J kg}^{-1}\text{ K}^{-1}

Specific heat capacity measures the heat capacity per unit mass of the material.

3

Compare the computed values

Highest specific heat capacity: Block

P

(

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

); Highest total heat capacity: Block

Q

(

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

).

Block

P

requires the most heat per kilogram per Kelvin, while Block

Q

requires the most heat overall to raise its temperature by

1\text{ K}

due to its larger mass.

Key Concept

Distinction between Heat Capacity (

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

) and Specific Heat Capacity (

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

)

Estimated Time:1m 30s

Question 130Question →

A copper calorimeter of mass $0.15\text{ kg}$ contains $0.25\text{ kg}$ of water at $20.0^\circ\text{C}$ . Dry steam at $100.0^\circ\text{C}$ is passed into the water until the final temperature of the mixture reaches $40.0^\circ\text{C}$ . Assuming no heat is lost to the surroundings, what is the mass of steam condensed, in grams?

(Take specific heat capacity of water $= 4200\text{ J kg}^{-1}\text{ K}^{-1}$ , specific heat capacity of copper $= 400\text{ J kg}^{-1}\text{ K}^{-1}$ , and specific latent heat of vaporization of water $= 2.26 \times 10^6\text{ J kg}^{-1}$ )

Show answer & explanation

Answer: 8.84

Answer

8.84 g

By energy conservation, the heat gained by the cold water and copper vessel must equal the total heat released by the condensing steam and the cooling of that condensed water. The heat gained is (0.25 kg × 4200 J/kg·K + 0.15 kg × 400 J/kg·K) × 20.0 K = 22,200 J. The heat lost per kilogram of steam is 2,260,000 J/kg + 4200 J/kg·K × 60.0 K = 2,512,000 J/kg. Dividing 22,200 J by 2,512,000 J/kg gives 0.0088376 kg, which corresponds to 8.84 g.

Step-by-Step Solution

1

Calculate the thermal energy absorbed by the cold water and copper calorimeter

Q_gained = 22,200 J

Both the water and copper calorimeter increase in temperature from 20.0°C to 40.0°C (ΔT = 20.0 K).

2

Express the heat released by mass m_s of steam as it condenses and cools to 40.0°C

Q_lost = m_s * 2,512,000 J

Steam releases latent heat when condensing at 100.0°C (m_s * L_v) and sensible heat when the condensed water cools from 100.0°C to 40.0°C (m_s * c_w * 60.0).

3

Apply the principle of conservation of thermal energy and solve for mass m_s in grams

m_s = 8.84 g

Setting Q_gained equal to Q_lost yields m_s = 22,200 / 2,512,000 = 0.0088376 kg, which equals 8.84 g.

Key Concept

Thermal energy balance involving phase change (latent heat of vaporization) and sensible heat exchange

Question 131Question →

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 132Question →

The specific latent heat of vaporization of a pure substance is substantially greater than its specific latent heat of fusion because vaporizing requires completely separating the molecules against intermolecular forces and performing work against external pressure, whereas melting requires only overcoming the long-range order of the crystalline lattice.

True

False

Show answer & explanation

Answer: True

Answer

True. The specific latent heat of vaporization is substantially larger than the specific latent heat of fusion because vaporization requires completely overcoming intermolecular cohesive forces to transition molecules into the gas phase, in addition to performing work against atmospheric pressure during expansion, whereas melting only disrupts long-range lattice order while keeping molecular spacing nearly identical.

The statement is true because vaporization involves breaking all intermolecular bonds to create a gas phase and doing external work against atmospheric pressure during large volume expansion, whereas fusion only weakens lattice alignment without significant separation or expansion.

Step-by-Step Solution

1

Analyze energy and structural changes during fusion (melting).

Melting breaks long-range crystalline order, but molecules remain in close contact in the liquid state with minimal volume change.

Only a small fraction of intermolecular potential energy is changed, and negligible work is done against external pressure.

2

Analyze energy and structural changes during vaporization (boiling).

Vaporization requires complete separation of molecules against attractive forces to form a gas, accompanied by a large increase in volume.

Energy must be supplied both to overcome intermolecular attractions completely and to perform mechanical work (

P\Delta V

) against surrounding atmospheric pressure.

3

Compare the thermodynamic energy requirements (

L_v

vs.

L_f

).

Because full molecular separation and expansion work require far more energy than minor lattice disruption,

L_v \gg L_f

.

This confirms the physics reasoning in the statement is true.

Key Concept

Microscopic and Thermodynamic Basis of Latent Heat of Fusion vs. Vaporization

Question 133Question →

A sample of ice with a mass of $0.20\text{ kg}$ at $0^\circ\text{C}$ absorbs $67,200\text{ J}$ of thermal energy to melt completely into water at $0^\circ\text{C}$ . What is the specific latent heat of fusion of ice in $\text{J kg}^{-1}$ ?

Show answer & explanation

Answer: 336000

Answer

The specific latent heat of fusion of ice is

336,000\text{ J kg}^{-1}

.

For a change of state from solid to liquid at constant temperature, the thermal energy transferred is related to mass by

Q = m L_f

. Dividing the absorbed energy (

67,200\text{ J}

) by the mass (

0.20\text{ kg}

) yields the specific latent heat of fusion,

336,000\text{ J kg}^{-1}

.

Step-by-Step Solution

1

Identify the given values from the problem statement

Heat energy

Q = 67,200\text{ J}

, Mass

m = 0.20\text{ kg}

These parameters are required to calculate the specific latent heat.

2

Apply the energy formula for phase transition at constant temperature

Q = m L_f

During melting, temperature remains constant at

0^\circ\text{C}

, so heat absorbed depends only on mass and specific latent heat of fusion.

3

Solve for the specific latent heat of fusion

L_f

L_f = \frac{Q}{m} = \frac{67,200}{0.20} = 336,000\text{ J kg}^{-1}

Dividing total energy by mass gives heat required per kilogram.

Key Concept

Specific Latent Heat of Fusion

Question 134Question →

A solid sample of lead with a mass of $0.80\text{ kg}$ is kept at its melting point of $327^\circ\text{C}$ . If $15,000\text{ J}$ of thermal energy is supplied to the sample, calculate the mass of lead, in kilograms, that remains in the solid state. (Take the specific latent heat of fusion of lead as $2.5 \times 10^4\text{ J/kg}$ ).

Show answer & explanation

Answer: 0.2

Answer

The mass of lead remaining in the solid state is

0.20\text{ kg}

.

Thermal energy

Q = 15,000\text{ J}

supplied to lead at its melting point melts a portion calculated by

m_{\text{melted}} = \frac{Q}{L_f} = \frac{15,000}{25,000} = 0.60\text{ kg}

. Subtracting this melted mass from the original

0.80\text{ kg}

yields

0.20\text{ kg}

of remaining solid lead.

Step-by-Step Solution

1

Calculate the mass of lead that melts.

Melted mass

m_{\text{melted}} = 0.60\text{ kg}

.

At the melting point, thermal energy supplied goes entirely into phase change without changing temperature:

Q = m_{\text{melted}} L_f

.

2

Determine the remaining mass of solid lead.

Remaining solid mass

m_{\text{solid}} = 0.20\text{ kg}

.

The un-melted portion equals the initial total mass minus the mass that has melted (

m_{\text{solid}} = m_{\text{total}} - m_{\text{melted}}

).

Key Concept

Latent Heat of Fusion and Phase Change

Question 135Question →

An electric immersion heater rated at $500\text{ W}$ is used to heat a liquid of mass $0.40\text{ kg}$ from an initial temperature of $25^\circ\text{C}$ to its boiling point at $75^\circ\text{C}$ , and then vaporize half of the liquid at constant temperature. If the specific heat capacity of the liquid is $2500\text{ J kg}^{-1}\text{ K}^{-1}$ and its specific latent heat of vaporization is $5.0 \times 10^5\text{ J kg}^{-1}$ , what is the total time required for this process?

300\text{ s}

500\text{ s}

200\text{ s}

10,100\text{ s}

Show answer & explanation

Answer: $300\text{ s}$

Answer

The total time required for the process is

300\text{ s}

.

The correct response combines the energy needed for thermal elevation to the boiling point (

50,000\text{ J}

) with the energy required for vaporizing half the mass at constant temperature (

100,000\text{ J}

), yielding a total energy of

150,000\text{ J}

. Dividing this total energy by the

500\text{ W}

power rating yields exactly

300\text{ s}

.

Step-by-Step Solution

1

Calculate the sensible heat (

Q_1

) required to heat

0.40\text{ kg}

of liquid from

25^\circ\text{C}

to

75^\circ\text{C}

.

Q_1 = m c \Delta T = 0.40\text{ kg} \times 2500\text{ J kg}^{-1}\text{ K}^{-1} \times (75 - 25)\text{ K} = 50,000\text{ J}

.

Before phase change can occur, the liquid must first reach its boiling point.

2

Calculate the latent heat (

Q_2

) required to vaporize half of the liquid mass (

m_{\text{vap}} = 0.20\text{ kg}

).

Q_2 = m_{\text{vap}} L_v = 0.20\text{ kg} \times 5.0 \times 10^5\text{ J kg}^{-1} = 100,000\text{ J}

.

Phase change at the boiling point occurs at constant temperature and depends strictly on latent heat.

3

Sum the total heat energy (

Q_{\text{total}}

) and compute time (

t

) using heater power (

P = 500\text{ W}

).

Q_{\text{total}} = Q_1 + Q_2 = 150,000\text{ J}

, so

t = \frac{Q_{\text{total}}}{P} = \frac{150,000\text{ J}}{500\text{ W}} = 300\text{ s}

.

Thermal power is defined as energy supplied per unit time (

P = \frac{Q}{t}

).

Key Concept

Latent Heat and Energy Balance in Phase Transitions

Estimated Time:2m 30s

Question 136Question →

If two solid blocks are manufactured from the exact same uniform metal but one has twice the mass of the other, the block with the larger mass will have a higher heat capacity, even though both blocks have the same specific heat capacity.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because specific heat capacity is an intensive property depending solely on the material, while heat capacity is an extensive property proportional to mass.

Specific heat capacity (

c

) is a characteristic property of a substance (

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

) and is independent of sample size. Heat capacity (

C = mc

) measures total thermal energy required per unit temperature rise (

\text{J K}^{-1}

) and scales directly with mass. Therefore, a larger mass of the same metal has a higher heat capacity while maintaining an identical specific heat capacity.

Step-by-Step Solution

1

Define specific heat capacity and identify its nature.

Specific heat capacity (

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

) is an intensive property, meaning it remains constant for any sample of the same material regardless of mass.

Intensive properties do not depend on the quantity of matter present.

2

Define heat capacity and relate it to mass and specific heat capacity.

Heat capacity (

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

) is an extensive property and is directly proportional to mass (

m

).

Extensive properties scale with the size or mass of the system.

3

Compare the two blocks.

Since both blocks are made of the same metal,

c

is identical for both. However, because

C = mc

, the block with greater mass has a higher heat capacity.

Doubling mass doubles total heat capacity while leaving specific heat capacity unchanged.

Key Concept

Distinction between Intensive (Specific Heat Capacity) and Extensive (Heat Capacity) Thermal Properties

Question 137Question →

On a warm day, the saturated vapour pressure of water at an ambient temperature of $25^\circ\text{C}$ is $24\text{ mmHg}$ . If the actual partial pressure of water vapour in the air at this temperature is $18\text{ mmHg}$ , what is the relative humidity of the atmosphere?

75\%

133\%

25\%

43\%

Show answer & explanation

Answer: $75\%$

Answer

The relative humidity of the atmosphere is

75\%

.

The relative humidity is defined as the ratio of the partial vapour pressure of water present in the air to the saturated vapour pressure at the ambient temperature, expressed as a percentage. Dividing

18\text{ mmHg}

by

24\text{ mmHg}

yields

0.75

, which equals

75\%

.

Step-by-Step Solution

1

Identify the formula for relative humidity.

\text{Relative Humidity} = \left(\frac{\text{Partial Vapour Pressure}}{\text{Saturated Vapour Pressure at same temp}}\right) \times 100\%

Relative humidity measures how close atmospheric air is to saturation at a given temperature.

2

Substitute the given values into the formula.

\text{Relative Humidity} = \left(\frac{18\text{ mmHg}}{24\text{ mmHg}}\right) \times 100\%

The partial pressure is

18\text{ mmHg}

and the saturated vapour pressure is

24\text{ mmHg}

.

3

Calculate the final percentage.

\text{Relative Humidity} = 0.75 \times 100\% = 75\%

Simplifying

\frac{18}{24}

gives

\frac{3}{4}

, which corresponds to

0.75

or

75\%

.

Key Concept

Relative humidity is the ratio of the actual mass (or partial pressure) of water vapour in a given volume of air to the maximum mass (or saturated vapour pressure) of water vapour that the same volume of air can hold at that same temperature, expressed as a percentage.

Question 138Question →

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 139Question →

An ice block of mass $0.15\text{ kg}$ at $-10^\circ\text{C}$ absorbs thermal energy until it turns completely into liquid water at $0^\circ\text{C}$ . Taking the specific heat capacity of ice as $2100\text{ J kg}^{-1}\text{ K}^{-1}$ and the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J kg}^{-1}$ , what is the total thermal energy supplied in joules?

Show answer & explanation

Answer: 53550

Answer

The total thermal energy supplied is 53,550 J.

To transform sub-zero ice into liquid water at its melting point, thermal energy is absorbed in two distinct stages: warming the ice from

-10^\circ\text{C}

to

0^\circ\text{C}

(

Q_1 = 0.15 \times 2100 \times 10 = 3150\text{ J}

) and melting the ice completely at

0^\circ\text{C}

(

Q_2 = 0.15 \times 3.36 \times 10^5 = 50400\text{ J}

). Adding these yields a total energy of

53550\text{ J}

.

Step-by-Step Solution

1

Calculate the thermal energy required to warm the solid ice from

-10^\circ\text{C}

to its melting point (

0^\circ\text{C}

).

Q_1 = 3150\text{ J}

Before the phase transition can begin, sensible heat must be supplied to raise the ice temperature using

Q_1 = m c_{\text{ice}} \Delta T

.

2

Calculate the thermal energy required to melt the ice at constant temperature (

0^\circ\text{C}

).

Q_2 = 50400\text{ J}

Phase transformation at constant temperature requires latent heat of fusion using

Q_2 = m L_f

.

3

Sum the energy required for both stages to determine total energy.

Q_{\text{total}} = 53550\text{ J}

The total energy supplied is the sum of sensible warming heat (

Q_1

) and latent melting heat (

Q_2

).

Key Concept

Multi-step heat energy calculation involving sensible heat (

m c \Delta T

) and latent heat of fusion (

m L_f

).

Question 140Question →

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

Thermal Physics

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