**IBDP Physics HL 2025 – B.1 Thermal energy transfers HL Paper 1**** Exam Style Questions **

## IBDP Physics 2025 HL Paper 1 – All Chapters

## B.1 Thermal energy transfers HL Paper 1

Kinetic Theory, Temperature, Internal Energy, Specific Heat Capacity, Latent Heat, Heat Transfer, Stefan-Boltzmann Law

### Question- B.1 Thermal energy transfers

The temperature of an object is changed from \(\theta_1{ }^{\circ} \mathrm{C}\) to \(\theta_2{ }^{\circ} \mathrm{C}\). What is the change in temperature measured in kelvin?

A. \(\left(\theta_2-\theta_1\right)\)

B. \(\left(\theta_2-\theta_1\right)+273\)

C. \(\left(\theta_2-\theta_1\right)-273\)

D. \(273-\left(\theta_2-\theta_1\right)\)

**▶️Answer/Explanation**

Ans:A

The change in temperature measured in kelvin is equal to the change in temperature in degrees Celsius because the Kelvin scale is based on the Celsius scale, but it starts at absolute zero (0 K = -273.15°C). Therefore, the correct answer is: \(\left(\theta_2 – \theta_1\right)\)

### Question

A metal cube \(X\) of length \(L\) is heated gaining thermal energy \(Q\). Its temperature rises by \(\Delta T\). \(A\) second cube \(Y\), of length \(2 L\), made of the same material, gains thermal energy of \(2 Q\).

What is the temperature rise of \(\mathrm{Y}\) ?

A. \(\frac{\Delta T}{8}\)

B. \(\frac{\Delta T}{4}\)

C. \(\Delta T\)

D. \(2 \Delta T\)

**▶️Answer/Explanation**

Ans:B

The change in thermal energy (\(Q\)) is related to the mass of the material, the specific heat capacity (\(c\)), and the change in temperature (\(\Delta T\)) by the equation:

\[Q = mc\Delta T\]

For both cubes \(X\) and \(Y\), made of the same material, the specific heat capacity is the same (\(c\)).

Now, let’s consider cube \(X\). Its length is \(L\), so its volume (\(V_X\)) is \(L^3\). Cube \(Y\) has twice the length (\(2L\), so its volume (\(V_Y\)) is \((2L)^3 = 8L^3\).

Since the mass (\(m\)) is directly proportional to volume and density, and both cubes are made of the same material, we can say that:

\[m_X = k \cdot V_X\]

\[m_Y = k \cdot V_Y\]

Where \(k\) is a constant.

So, the ratio of the mass of cube \(Y\) to the mass of cube \(X\) is:

\[\frac{m_Y}{m_X} = \frac{k \cdot V_Y}{k \cdot V_X} = \frac{8L^3}{L^3} = 8\]

Now, let’s consider the thermal energy gained:

For cube \(X\): \(Q_X = m_Xc\Delta T\)

For cube \(Y\): \(Q_Y = m_Yc\Delta T_Y\)

Given that \(Q_Y = 2Q_X\) and \(\frac{m_Y}{m_X} = 8\), we can write:

\[2Q_X = 8(m_Xc\Delta T)\]

Now, we can simplify:

\[2Q_X = 8Q_X\]

Divide both sides by 2:

\[Q_X = 4Q_X\]

This implies that \(\Delta T_Y\) for cube \(Y\) must be four times the change in temperature \(\Delta T\) for cube \(X\).

So,

\[\Delta T_Y = 4\Delta T\]

The correct answer is B. \(\frac{\Delta T}{4}\).

An ideal gas expands at constant pressure. The graph shows the relationship between pressure *P* and volume *V* for this change.

The change in the internal energy of the gas during this expansion is 1800 J. What is the amount and the direction of thermal energy transferred?

A. 3000 J into the gas

B. 3000 J out of the gas

C. 600 J into the gas

D. 600 J out of the gas.

**Answer/Explanation**

Answer – A

Work \(= – ve \) ( during expansion )

\( ΔU = Q – W \)

Work done \(= P\Delta V\)

\(=4\times 10^{5} (4- 1)10^{-3}\)

=1200 J

\(ΔU = Q – W \)

\( ΔU = 1800- (-1200)\)

\( ΔU = +3000 J \)

Water at a temperature of 0 °C is kept in a thermally insulated container. A lump of ice, also at 0 °C, is placed in the water and completely submerged.

Which of the following is true in respect of both the net amount of ice that will melt and the change in temperature of the water?

**Answer/Explanation**

Ans: C

when o°c lump of ice placed at a temperature 0°c water which is kept thermally insulated container, at first

Some ice melt into water because the latent heat present in water more than ice but due to insulated container and for me melting of ice the required amount of heat cannot supply from Surrounding thus the same

amount of water is also converted into ice, so there is mo charge due to equilibrium condition. So the the answer is none will be heat and no change in temperature

Which of the following correctly identifies the properties of the molecules of a substance that determine the substance’s internal energy?

A. The total potential energy and random kinetic energy

B. The random kinetic energy

C. The total gravitational potential energy and random kinetic energy

D. The total potential energy

**Answer/Explanation**

Answer -A

Internal energy is the kinetic and potential energy associated with the random motion of the molecules of an object.