IB Physics HL B 1 Thermal Energy Transfers Questions

Explore how particle models explain temperature and internal energy, how thermal energy moves or changes state, and how radiation links to stellar brightness and temperature.

Exam points

Relate particle arrangement to state and density; define internal energy and link Kelvin temperature to mean kinetic energy.
Identify net heat-transfer direction and explain kinetic- or potential-energy changes during phase changes.
Use Q=mcΔT and Q=mL for heating, cooling and phase changes, including staged energy and power calculations.
Explain conduction, convection and radiation; calculate conduction rate from conductivity, area, thickness and ΔT.
Use black-body power and b=L/(4πd²) to relate luminosity, apparent brightness and distance.
Apply Wien’s law, λmaxT=2.9×10−3 m K, to infer temperature or peak wavelength.

Covered syllabus points

B.1.1—Molecular states B.1.2—Density B.1.3—Temperature scales B.1.4—Temperature scale changes B.1.5—Kelvin temperature and kinetic energy B.1.6—Internal energy B.1.7—Thermal transfer direction B.1.8—Phase change B.1.9—Specific heat and latent heat B.1.10—Thermal transfer mechanisms B.1.11—Conduction B.1.12—Conduction rate B.1.13—Convection B.1.14—Black-body radiation B.1.15—Apparent brightness B.1.16—Luminosity and brightness B.1.17—Wien’s displacement law

[Maximum number: 9]

Ceres is a dwarf planet in the asteroid belt. The following data are available.

\begin{array}{ll} \text { Mean distance of Ceres from the Sun } & =4.4 \times 10^{11} \mathrm{~m} \\ \text { Mean power output of the Sun } & =3.8 \times 10^{26} \mathrm{~W} \end{array}

(a)

Determine the mean temperature of Ceres assuming that it acts as a black-body radiator.

[ 3 ]

average intensity at Ceres orbit $=\left(\frac{3.8 \times 10^{26}}{4 \times \pi \times\left(4.4 \times 10^{11}\right)^{2}}\right) \lessdot=156 \mathrm{Wm}^{-2} \| \vee$ average incident intensity $=\frac{156}{4}=\ll 39 \gg$

temperature $<=\left(\frac{39}{5.67 \times 10^{-8}}\right)^{\frac{1}{4}}>=160 \ll \mathrm{~K} \gg$

Marking guidance:

Allow ECF from MP1 and MP2

(b)

Ceres has a solid rocky core covered with solid ice. The mean temperature is higher than your answer in (a)(i) because radioactive nuclei in the centre of Ceres are decaying. Outline how the energy from the radioactive decay reaches the surface.

[ 2 ]

(c)

At low temperatures such as the mean temperature of Ceres, water undergoes a phase change directly from solid to gas.

[ 4 ]

(i)

Compare the molecular conditions of the solid phase and the gas phase at the same temperature.

[ 3 ]

(ii)

The maximum surface temperature of Ceres is $-38^{\circ} \mathrm{C}$ . Observations show that significant quantities of water vapour are released from the surface of Ceres every second when the temperature is at this maximum. Calculate the mean kinetic energy of a molecule of water vapour at this temperature.

[ 1 ]

4.9\times10^{-21}\,\mathrm{J}

[Maximum number: 3]

(a)

A cylindrical spacecraft of cross-sectional area S moves with velocity v. The spacecraft enters a region of dust of density $\rho$ . All the dust that comes into contact with the forward cross-sectional area of the spacecraft sticks to the spacecraft increasing its mass.

[ 1 ]

(i)

Show that in time $\Delta t$ the mass of the spacecraft will increase by an amount $\rho \operatorname{Sv} \Delta t$ .

[ 1 ]

Identifies length of cylinder as $v \Delta t$ so volume as $S \times v \times \Delta t$ to show the result

(b)

The graph shows the variation with time of the height of the International Space Station (ISS) during a 30-day period.

The following data are available:

\begin{aligned} \text { Mass of Earth } & =5.98 \times 10^{24} \mathrm{~kg} \\ \text { Mass of the ISS } & =4.20 \times 10^{5} \mathrm{~kg} \\ \text { Radius of the Earth } & =6.38 \times 10^{6} \mathrm{~m} \end{aligned}

For the 30-day period indicated in the graph the loss of total energy of the ISS is $4.5 \times 10^{9} \mathrm{~J}$ .

[ 2 ]

(i)

Estimate the increase in the temperature of the ISS assuming all the lost energy went into thermal energy of the ISS. Take the specific heat capacity of the ISS to be $500 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ .

[ 2 ]

$\begin{aligned} \Delta T=\frac{Q}{m c} \quad \Delta T=\text { « } \frac{4.5 \times 10^{9}}{4.2 \times 10^{5} \times 500}=\gg 21 \mathrm{~K} \text { OR } 21^{\circ} \mathrm{C} \quad \end{aligned}$

Award [2] if 21 is seen as the answer without working

[Maximum number: 1]

The temperature of an object is $-153^{\circ} \mathrm{C}$ . Its temperature is raised to $273^{\circ} \mathrm{C}$ . What is the temperature change of the object?

699 K

426 K

153 K

120 K

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question 2

Question (a)

Question (i)

Question (b)

Question (i)

Question 3