IBDP Physics SL 2025 -B.1 Thermal energy transfers SL Paper 1 Exam Style Questions
IBDP Physics SL Paper 1 – All Chapters
Topic: B.1 Thermal energy transfers SL Paper 1
Kinetic Theory, Temperature, Internal Energy, Specific Heat Capacity, Latent Heat, Heat Transfer, Stefan-Boltzmann Law
Question-B.1 Thermal energy transfers SL Paper 1
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
The temperature of a gas increases from \(100 \mathrm{~K}\) to \(330 \mathrm{~K}\). What is the change in temperature of the gas in degrees Celsius?
A. 503
B. 230
C. -43
D. -230
▶️Answer/Explanation
Ans:B
To find the change in temperature in degrees Celsius, you can use the following relationship between the Kelvin (K) and Celsius (°C) scales:
\(\Delta T (\text{°C}) = \Delta T (\text{K})\)
The change in temperature in degrees Celsius is equal to the change in temperature in Kelvin. So, in this case:
\(\Delta T (\text{°C}) = 330 \, \text{K} – 100 \, \text{K} = 230 \, \text{K}\)
So, the change in temperature of the gas in degrees Celsius is 230°C.
Question
The total potential energy and random kinetic energy of the molecules of an object is equal to the
A. heat energy in the object.
B. internal energy of the object.
C. thermal energy in the object.
D. work stored in the object.
Answer/Explanation
Markscheme
B
Internal energy is the sum of total Kinetic energy ( translational + rotational ) of its molecules and the total potential energy of its molecules with respect to each other
Question
A temperature of 23 K is equivalent to a temperature of
A. – 300 °C.
B. – 250 °C
C. + 250 °C
D. + 300 °C.
Answer/Explanation
Markscheme
B
\(T(k) = 273.15 + T(^{\circ}c)\)
\(23K = 273.15 + T(^{\circ}c)\)
\(T(^{\circ}c)=-250^{\circ}c\)
Question
A substance is heated at constant power. The graph shows how the temperature T of the substance varies with time t as the state of the substance changes from liquid to gas.
What can be determined from the graph?
A. The specific heat capacity of the gas is smaller than the specific heat capacity of the liquid.
B. The specific heat capacity of the gas is larger than the specific heat capacity of the liquid.
C. The specific latent heat of fusion of the substance is less than its specific latent heat of vaporization.
D. The specific latent heat of fusion of the substance is larger than its specific latent heat of vaporization.
Answer/Explanation
Markscheme
B
Slope of OA >Slope of BC
\(\Rightarrow (\frac{dT}{dQ})_{OA}=(\frac{dT}{dQ})_{BC}\)
\(\Rightarrow \frac{1}{mS_{liquid}}=\frac{1}{mS_{gas}}\)
\(\Rightarrow S_{gas}>S_{liquid}\)
Thus, the specific heat capacity of the gas is larger than the specific heat capacity of the liquid.