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IBDP Physics- B.1 Thermal energy transfers- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

Insulated solid copper rods of different diameters and fixed length are placed in thermal contact with two objects X and Y, maintained at different temperatures \(T_x\) and \(T_y\) respectively. The diagram shows the setup with one such rod.

Which graph shows the variation with rod diameter \(d\) of the rate of thermal energy transfer \(\dfrac{\Delta Q}{\Delta t}\) along the rod?

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

The rate of thermal energy transfer by conduction is given by

\(\dfrac{\Delta Q}{\Delta t}=\dfrac{kA\Delta T}{L}\)

where \(k\) is the thermal conductivity, \(A\) is the cross-sectional area, \(\Delta T\) is the temperature difference and \(L\) is the length of the rod.

For a cylindrical rod, the cross-sectional area is

\(A=\pi r^2\)

Since the diameter is \(d=2r\),

\(A=\pi\left(\dfrac{d}{2}\right)^2=\dfrac{\pi d^2}{4}\)

Therefore,

\(\dfrac{\Delta Q}{\Delta t}=\dfrac{k\pi d^2\Delta T}{4L}\)

Hence, the rate of thermal energy transfer is proportional to the square of the diameter:

\(\dfrac{\Delta Q}{\Delta t}\propto d^2\)

A \(d^2\) relationship produces a curve that becomes progressively steeper as \(d\) increases.

Thus, the correct graph is \( \boxed{\mathrm{D}} \).

Hence, the correct answer is \( \boxed{\mathrm{D}} \).

Question 

\(80\,\mathrm{g}\) of a liquid is heated with a constant power output of \(100\,\mathrm{W}\). All the power goes into the liquid. After \(60\,\mathrm{s}\), the rise in temperature is \(50\,\mathrm{K}\).

What is the specific heat capacity of the liquid in \(\mathrm{J\,kg^{-1}\,K^{-1}}\)?

(A) \(120\)
(B) \(960\)
(C) \(1200\)
(D) \(1500\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

The thermal energy supplied is

\(Q=Pt\)

\(Q=(100)(60)=6000\,\mathrm{J}\)

The mass is

\(m=80\,\mathrm{g}=0.080\,\mathrm{kg}\)

Using the specific heat capacity equation

\(Q=mc\Delta T\)

Therefore,

\(c=\dfrac{Q}{m\Delta T}\)

\(c=\dfrac{6000}{(0.080)(50)}\)

\(c=1500\,\mathrm{J\,kg^{-1}\,K^{-1}}\)

Hence, the correct answer is \( \boxed{\mathrm{D}} \).

Question

A metal cube \(X\) of side length \(L\) is heated and gains thermal energy \(Q\). Its temperature increases by \(\Delta T\).

A second cube \(Y\), made of the same material and having side length \(2L\), gains thermal energy \(2Q\).

What is the temperature rise of cube \(Y\)?
(A) \( \dfrac{\Delta T}{8} \)
(B) \( \dfrac{\Delta T}{4} \)
(C) \( \Delta T \)
(D) \( 2\Delta T \)
▶️ Answer/Explanation
Detailed solution

The thermal energy gained by a body is given by \( Q = mc\Delta T \), where \(m\) is the mass and \(c\) is the specific heat capacity.

Since both cubes are made of the same material, \(c\) is the same for both. The mass of each cube is proportional to its volume.

For cube \(X\): volume \(= L^3\).
For cube \(Y\): volume \(= (2L)^3 = 8L^3\).
Hence, \( m_Y = 8m_X \).

For cube \(X\): \( Q = m_X c \Delta T \).
For cube \(Y\): \( 2Q = m_Y c \Delta T_Y \).

Substituting \( m_Y = 8m_X \):
\( 2(m_X c \Delta T) = 8m_X c \Delta T_Y \).

Cancelling common terms gives \( 2\Delta T = 8\Delta T_Y \), so \( \Delta T_Y = \dfrac{\Delta T}{4} \).

✅ Answer: (B)

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