IB Physics Quiz: Understand Thermal Energy Transfers

20 questions · exam conditions

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Understand Thermal Energy TransfersQuestion 1 of 20

The average kinetic energy of the molecules in a certain mass of an ideal gas is $E_k$ . The absolute temperature of the gas is doubled, and half of the gas molecules are removed. What is the new average kinetic energy of the remaining molecules?

E_k / 2

E_k

2 E_k

4 E_k

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IB Physics Quiz

IB Physics Quiz: Understand Thermal Energy Transfers

Practice Understand Thermal Energy Transfers in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Thermal Energy Transfers, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

$E_k / 2$
$E_k$
$2 E_k$ (correct answer)
$4 E_k$

Explanation: For an ideal gas, the average kinetic energy of the molecules is directly proportional to the absolute temperature (in Kelvin), given by the relationship

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

. The average kinetic energy depends only on temperature, not on the number of molecules or the total mass of the gas. Since the absolute temperature is doubled, the average kinetic energy of the molecules also doubles to

2 E_k

Question 2

An electric kettle with a power rating of 2.0 kW is used to boil water. After the water reaches 100 °C, it is observed that 150 g of water turns into steam in 3.0 minutes. What is the specific latent heat of vaporization of water calculated from this data?

$2.4 \times 10^3$ J kg⁻¹
$4.0 \times 10^4$ J kg⁻¹
$2.4 \times 10^6$ J kg⁻¹ (correct answer)
$3.6 \times 10^8$ J kg⁻¹

Explanation: First, calculate the total energy

Q

supplied by the kettle. Power

P = 2.0

kW = 2000 W. Time

t = 3.0

min = 180 s. Energy

Q = P \times t = 2000 \times 180 = 360000

J. This energy vaporized a mass

m = 150

g = 0.150 kg. The specific latent heat of vaporization

L_v

is given by

Q = mL_v

. Therefore,

L_v = Q/m = 360000 / 0.150 = 2400000

J kg⁻¹, which is

2.4 \times 10^6

J kg⁻¹.

Question 3

Specific heat capacity of water = $4.2 \times 10^3$ J kg⁻¹ K⁻¹ Specific latent heat of fusion of ice = $3.3 \times 10^5$ J kg⁻¹

A 50 g ice cube at 0 °C is added to 200 g of water at 25.9 °C in a perfectly insulated container. What is the final equilibrium temperature of the mixture?

0.0 °C
5.0 °C (correct answer)
6.3 °C
20.7 °C

Explanation: First, check if all the ice melts. The heat required to melt the ice is

Q_{melt} = m_{ice}L_f = (0.050)(3.3 \times 10^5) = 16500

J. The maximum heat the water can lose by cooling to 0 °C is

Q_{cool} = m_{water}c_w\Delta T = (0.200)(4200)(25.9 - 0) = 21756

J. Since

Q_{cool} > Q_{melt}

, all the ice melts and the final temperature

T_f

is above 0 °C. By conservation of energy: Heat lost by water = Heat to melt ice + Heat to warm melted ice.

m_w c_w (25.9 - T_f) = m_{ice} L_f + m_{ice} c_w (T_f - 0)

(0.200)(4200)(25.9 - T_f) = 16500 + (0.050)(4200)(T_f)

840(25.9 - T_f) = 16500 + 210T_f

21756 - 840T_f = 16500 + 210T_f

5256 = 1050T_f

T_f = 5.0

°C.

Question 4

A metal spoon and a wooden spoon are both at room temperature. The metal spoon feels colder to the touch than the wooden spoon. Which statement best explains this observation?

The metal has a lower specific heat capacity, so it contains less thermal energy.
The metal is a better thermal conductor, so it transfers heat from the hand faster. (correct answer)
The wood is a better thermal insulator, so it transfers cold to the hand slower.
The metal has a lower temperature because it radiates heat away more effectively.

Explanation: Both spoons are at the same room temperature. The sensation of 'cold' is due to the rate of heat transfer away from your hand. Metal is a much better thermal conductor than wood. When you touch the metal spoon, it conducts heat away from your hand rapidly, creating a cold sensation. Wood, being an insulator, conducts heat away much more slowly.

Question 5

Two stars, X and Y, radiate as black bodies. The radius of star X is one-quarter the radius of star Y ( $R_X = 0.25 R_Y$ ). The total power radiated by star X is 16 times the power radiated by star Y ( $L_X = 16 L_Y$ ). What is the ratio of the surface temperature of X to that of Y ( $T_X / T_Y$ )?

2
4 (correct answer)
8
16

Explanation: The Stefan-Boltzmann law states that

L = \sigma A T^4

. For a sphere,

A = 4\pi R^2

, so

L = \sigma (4\pi R^2) T^4

. We can form a ratio:

L_X / L_Y = (R_X / R_Y)^2 (T_X / T_Y)^4

. We are given

L_X / L_Y = 16

and

R_X / R_Y = 0.25

1/4

. Substituting these values:

16 = (0.25)^2 (T_X / T_Y)^4 \Rightarrow 16 = (1/16) (T_X / T_Y)^4

. To solve for the temperature ratio, we rearrange:

(T_X / T_Y)^4 = 16 \times 16 = 256

. Therefore,

T_X / T_Y = \sqrt[4]{256} = 4

Question 6

Star Sirius has a peak emission wavelength $\lambda$ and a luminosity $L$ . Star Procyon has a peak emission wavelength of $1.5\lambda$ . Both stars can be modelled as black bodies and have approximately the same radius. What is the approximate luminosity of Procyon in terms of $L$ ?

0.20 L (correct answer)
0.44 L
2.25 L
5.06 L

Explanation: First, use Wien's law (

\lambda_{max} T = \text{constant}

) to find the temperature ratio.

T \propto 1/\lambda_{max}

, so

T_{Procyon} / T_{Sirius} = \lambda / (1.5\lambda) = 1/1.5 = 2/3

. Next, use the Stefan-Boltzmann law (

L = \sigma A T^4

). Since the radii are the same, their surface areas

A

are the same. Thus,

L \propto T^4

. The ratio of luminosities is

L_{Procyon} / L_{Sirius} = (T_{Procyon} / T_{Sirius})^4 = (2/3)^4 = 16/81 \approx 0.1975

. Therefore, the luminosity of Procyon is approximately 0.20

L

Question 7

A large mass of water in a lake at 2 °C is compared to a small cup of boiling water at 100 °C. Which statement is correct?

The lake water has a higher temperature and a greater internal energy.
The cup of water has a higher temperature and a greater internal energy.
The lake water has a lower temperature but a greater internal energy. (correct answer)
The cup of water has a lower temperature but a greater internal energy.

Explanation: Temperature is a measure of the average kinetic energy of the molecules. The boiling water (100 °C) is clearly at a higher temperature than the lake water (2 °C). Internal energy is the sum of the total kinetic and potential energies of all the molecules in the substance. Although the molecules in the lake have a lower average kinetic energy, the sheer number of molecules in the large mass of the lake is vastly greater than in the small cup. Consequently, the total internal energy of the lake water is much greater than that of the cup of boiling water.

Question 8

An object with emissivity $e$ is in an environment at absolute temperature $T_s$ . The object's absolute temperature is $T_o$ , and $T_o > T_s$ . The object radiates power at a rate of $P_{emit}$ and absorbs power at a rate of $P_{absorb}$ . What is the net rate of energy loss of the object?

$e \sigma A (T_o^4 - T_s^4)$ (correct answer)
$e \sigma A T_o^4$
$e \sigma A (T_o - T_s)^4$
$e \sigma A T_s^4$

Explanation: The rate of energy emission by the object is given by the Stefan-Boltzmann law for a non-black body:

P_{emit} = e \sigma A T_o^4

. The object also absorbs radiation from its surroundings. The power absorbed is determined by the temperature of the surroundings,

T_s

, so

P_{absorb} = e \sigma A T_s^4

. The net rate of energy loss is the difference between the power emitted and the power absorbed:

P_{net} = P_{emit} - P_{absorb} = e \sigma A T_o^4 - e \sigma A T_s^4 = e \sigma A (T_o^4 - T_s^4)

Question 9

A 200 g block of a metal with specific heat capacity $c$ absorbs 4000 J of thermal energy, and its temperature increases by 20 K. A 100 g block of a different metal with specific heat capacity $2c$ absorbs 3000 J of thermal energy. What is the temperature increase of the second block?

15 K (correct answer)
20 K
30 K
45 K

Explanation: This is a two-step problem. First, find the value of

c

for the first metal using

Q = mc\Delta T

4000 = (0.200)(c)(20)

, which gives

4000 = 4c

, so

c = 1000

J kg⁻¹ K⁻¹. The specific heat capacity of the second metal is

2c = 2000

J kg⁻¹ K⁻¹. Now, use

Q = mc\Delta T

for the second block:

3000 = (0.100)(2000)(\Delta T)

. This simplifies to

3000 = 200\Delta T

, so

\Delta T = 15

K. Alternatively, using ratios:

\Delta T_2 / \Delta T_1 = (Q_2/Q_1) \cdot (m_1/m_2) \cdot (c_1/c_2) = (3000/4000) \cdot (200/100) \cdot (c/2c) = (3/4) \cdot (2) \cdot (1/2) = 3/4

. So

\Delta T_2 = (3/4) \cdot 20 = 15

Question 10

A small, hot metal block is placed in contact with a large, cold metal block in a thermally isolated system. Which quantity must be equal for both blocks once the system has reached thermal equilibrium?

The magnitude of the change in internal energy of each block.
The final internal energy of each block.
The specific heat capacity of each block.
The final temperature of each block. (correct answer)

Explanation: Thermal equilibrium is defined as the state where there is no net flow of thermal energy between objects in thermal contact. This condition is met when the objects reach the same final temperature. While the heat lost by the hot block equals the heat gained by the cold block (so the magnitudes of their change in internal energy are equal), their final internal energies will be different due to their different masses and materials. Specific heat capacity is an intrinsic property of the material and does not change.

Question 11

The rate of heat flow by conduction through a solid bar is $P$ . The length of the bar is then doubled, and its cross-sectional area is halved. All other factors, including the temperature difference across its ends, remain constant. What is the new rate of heat flow?

$P / 4$ (correct answer)
$P / 2$
$P$
$4P$

Explanation: The rate of heat flow by conduction

P

is given by the formula

P = kA\Delta T / L

, where

k

is the thermal conductivity,

A

is the cross-sectional area,

\Delta T

is the temperature difference, and

L

is the length. The rate of flow is directly proportional to the area and inversely proportional to the length. If the length is doubled (

L \to 2L

) and the area is halved (

A \to A/2

), the new rate

P_{new}

will be proportional to

(A/2) / (2L) = A / (4L)

. This is one-quarter of the original proportionality, so the new rate of heat flow is

P/4

Question 12

A window pane is a single sheet of glass of thickness 5.0 mm and area 2.0 m². On a cold day, the inner surface is at 15 °C and the outer surface is at 5.0 °C. The thermal conductivity of the glass is 0.80 W m⁻¹ K⁻¹. What is the rate of heat loss through the window?

3.2 W
1.6 kW
3.2 kW (correct answer)
6.4 kW

Explanation: The rate of heat transfer by conduction is given by

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

. Here,

k = 0.80

W m⁻¹ K⁻¹,

A = 2.0

m²,

\Delta T = 15 - 5.0 = 10

K, and

\Delta x = 5.0

mm =

5.0 \times 10^{-3}

m. Substituting these values gives: Rate =

\frac{(0.80)(2.0)(10)}{5.0 \times 10^{-3}} = \frac{16}{5.0 \times 10^{-3}} = 3200

W. This is equal to 3.2 kW.

Question 13

A star radiates as a perfect black body with a surface temperature of 5800 K. The peak emission wavelength of a second star is observed to be half that of the first star. What is the surface temperature of the second star?

2900 K
4100 K
8200 K
11600 K (correct answer)

Explanation: Wien's displacement law states that the peak emission wavelength

\lambda_{max}

is inversely proportional to the absolute temperature

T

, i.e.,

\lambda_{max} T = \text{constant}

. Therefore,

\lambda_1 T_1 = \lambda_2 T_2

. We are given

T_1 = 5800

K and

\lambda_2 = 0.5 \lambda_1

. Substituting this into the equation gives

\lambda_1 (5800) = (0.5 \lambda_1) T_2

. The

\lambda_1

terms cancel, leaving

5800 = 0.5 T_2

. Solving for

T_2

gives

T_2 = 5800 / 0.5 = 11600

Question 14

Two identical rooms are separated by a wall. Room A is at $25°C$ and Room B is at $15°C$ . The wall has thermal conductivity $k$ and thickness $d$ . If the wall thickness is reduced to $d/2$ while simultaneously doubling the thermal conductivity to $2k$ , by what factor does the rate of heat transfer between the rooms change?

Increases by a factor of 2
Increases by a factor of 3
Increases by a factor of 4 (correct answer)
Increases by a factor of 6

Explanation: The rate of heat conduction is given by

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

. Initially:

\dot{Q}_1 = \frac{kA(10)}{d}

. After changes:

\dot{Q}_2 = \frac{(2k)A(10)}{d/2} = \frac{4kA(10)}{d}

. The ratio is

\frac{\dot{Q}_2}{\dot{Q}_1} = 4

. The thermal conductivity doubles (factor of 2) and the thickness halves (factor of 2), giving a combined factor of

2 \times 2 = 4

Question 15

Two thin metal plates are separated by a small air gap of width $d = 2\text{ mm}$ . The plates are at $60°C$ and $40°C$ respectively. Heat transfer occurs through conduction in the still air ( $k = 0.025 \text{ W/mK}$ ) and radiation between the plates ( $\epsilon = 0.8$ for both surfaces). Which mechanism dominates the heat transfer, and approximately what fraction of the total heat transfer does it represent?

Conduction dominates, representing approximately 85% of total heat transfer (correct answer)
Radiation dominates, representing approximately 75% of total heat transfer
Conduction dominates, representing approximately 65% of total heat transfer
Both mechanisms contribute equally, each representing approximately 50% of total heat transfer

Explanation: Conductive heat flux:

q_{cond} = k\frac{\Delta T}{d} = 0.025 \times \frac{20}{0.002} = 250 \text{ W/m}^2

. Radiative heat flux:

q_{rad} = \epsilon\sigma(T_1^4 - T_2^4) = 0.8 \times 5.67 \times 10^{-8} \times (333^4 - 313^4) \approx 0.8 \times 5.67 \times 10^{-8} \times 2.89 \times 10^{10} \approx 131 \text{ W/m}^2

. Total:

250 + 131 = 381 \text{ W/m}^2

. Conduction fraction:

\frac{250}{381} \approx 0.66

or about 66%. The closest answer is 65%, so conduction dominates with approximately 65% of the total.

Question 16

A window consists of two parallel glass panes separated by an air gap. In winter, the inside surface is at $20°C$ and the outside surface is at $-10°C$ . If the air gap width is doubled while keeping the glass thickness constant, which statement best describes the change in heat transfer?

Heat transfer rate decreases because the thermal resistance of the air gap increases proportionally with thickness
Heat transfer rate remains nearly constant because convection currents in the wider gap increase heat transfer (correct answer)
Heat transfer rate decreases significantly because radiation across the gap becomes the dominant mechanism
Heat transfer rate increases because the larger air volume stores more thermal energy

Explanation: While doubling the air gap thickness would decrease conductive heat transfer (

\dot{Q} \propto 1/d

), the wider gap allows for stronger convection currents to develop. In double-pane windows, convection in the air gap often dominates over pure conduction. The enhanced convection in the wider gap approximately compensates for the reduced conduction, keeping the overall heat transfer rate relatively constant. Option A ignores convection, C incorrectly emphasizes radiation which is minimal at these temperatures, and D confuses energy storage with energy transfer rate.

Question 17

A spherical object at $100°C$ is cooling in a room at $20°C$ primarily through convection. The convective heat transfer coefficient is $h = 25 \text{ W/m}^2\text{K}$ . If the sphere's radius is doubled while maintaining the same material and surface properties, how does the initial rate of cooling ( $\frac{dT}{dt}$ ) change?

Decreases by a factor of 2 because the surface area to volume ratio decreases (correct answer)
Increases by a factor of 2 because the surface area increases by a factor of 4
Decreases by a factor of 2 because the thermal mass increases by a factor of 8
Remains the same because both surface area and mass increase proportionally

Explanation: The rate of cooling depends on Newton's law of cooling:

\frac{dT}{dt} = -\frac{hA}{mc}(T-T_{ambient})

. For a sphere,

A = 4\pi r^2

and

m = \frac{4}{3}\pi r^3 \rho

. When radius doubles: new area =

4 \times

original, new mass =

8 \times

original. So

\frac{dT}{dt}_{new} = -\frac{h(4A)}{(8m)c}(T-T_{ambient}) = \frac{1}{2} \times \frac{dT}{dt}_{original}

. The cooling rate decreases by a factor of 2 because the surface area to volume ratio decreases.

Question 18

Which of the following is the most fundamental condition required for heat transfer by natural convection to occur within a fluid?

The fluid must be in a gravitational field. (correct answer)
The fluid's volume must change with pressure.
The fluid must be in contact with a solid surface.
The fluid must be a gas and not a liquid.

Explanation: Natural convection is driven by buoyancy forces that arise from density differences within the fluid. These density differences are typically caused by temperature variations. However, buoyancy force itself (the upward force exerted by a fluid that opposes the weight of an immersed object) only exists in the presence of a gravitational field. Without gravity, the less dense, warmer fluid would not rise, and the denser, cooler fluid would not sink, so no convection currents would form.

Question 19

A liquid in an open, insulated container is observed to cool as it evaporates. What is the primary reason for this cooling effect?

The ambient air pressure drops as vapor is created.
The molecules with the highest kinetic energy are the most likely to escape. (correct answer)
The remaining molecules do work on the escaping molecules, reducing their energy.
The formation of vapor requires an input of potential energy from the surroundings.

Explanation: The molecules in a liquid have a distribution of kinetic energies. Evaporation is the process where molecules at the surface with sufficient kinetic energy to overcome intermolecular forces escape into the vapor phase. The molecules that escape are, on average, the most energetic ones. This selective removal of high-energy molecules lowers the average kinetic energy of the molecules remaining in the liquid. Since temperature is a measure of the average kinetic energy, the liquid's temperature decreases.

Question 20

When a substance melts at a constant temperature, what are the changes in the average kinetic energy and the average potential energy of its molecules?

The average kinetic energy increases and the average potential energy remains constant.
The average kinetic energy remains constant and the average potential energy increases. (correct answer)
The average kinetic energy increases and the average potential energy increases.
The average kinetic energy remains constant and the average potential energy remains constant.

Explanation: Temperature is a measure of the average kinetic energy of the molecules. Since the substance melts at a constant temperature, the average kinetic energy of the molecules must remain constant. The energy added to the substance during melting (latent heat of fusion) is used to break the intermolecular bonds, increasing the separation between molecules. This increases the average potential energy of the molecules.