CourseCourseConceptConceptDocsDocuments

IB Physics SL B 2 Greenhouse Effect Questions

Practise IB Physics SL B.2 by building surface–atmosphere energy balances from solar input, albedo, thermal emission and greenhouse absorption.

Syllabus
First assessment 2025
Course
Physics SL
Level
SL

Exam points

  • calculate and interpret solar intensity, solar constant, S/4, albedo and the separation of reflected and absorbed energy
  • apply emissivity and Stefan–Boltzmann radiation to surface–atmosphere energy-balance diagrams and infrared emission
  • explain infrared absorption and re-emission, enhanced greenhouse forcing, albedo effects and evidence for human causes or mitigation

Covered syllabus points

B.2.1—Conservation of energyB.2.2—EmissivityB.2.3—AlbedoB.2.4—Earth’s albedo variationB.2.5—Solar constant SB.2.6—Mean solar intensityB.2.7—Greenhouse gasesB.2.8—Infrared absorptionB.2.9—Greenhouse modelsB.2.10—Enhanced greenhouse effect

Learn this topic

Study B.2 Greenhouse effect conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

Question 1

[Maximum number: 1]

In an energy-balance climate model, the power of the incoming radiation over an area A is PiP_{\mathrm{i}} and the power of the outgoing radiation over the same area is PoP_{\mathrm{o}}. The surface heat capacity is CsC_{\mathrm{s}}. What is the time taken to increase the temperature of the area by θ\theta ?

A

(Pi−Po)Csθ\frac{\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)}{C_{\mathrm{s}} \theta}

B

Csθ(Pi−Po)\frac{C_{\mathrm{s}} \theta}{\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)}

C

ACsθ(Pi−Po)\frac{A C_{\mathrm{s}} \theta}{\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)}

D

A(Pi−Po)Csθ\frac{A\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)}{C_{\mathrm{s}} \theta}

Question 2

[Maximum number: 8]

The diagram shows a simplified energy-balance model for the Earth surface–atmosphere system.

Figure for Question 2 — IB Physics SL

The following data are given:

 Average albedo of Earth =0.30 Average global temperature of the surface =288 K Average Earth-Sun distance =1.5×1011 m\begin{aligned} \text { Average albedo of Earth } & =0.30 \\ \text { Average global temperature of the surface } & =288 \mathrm{~K} \\ \text { Average Earth-Sun distance } & =1.5 \times 10^{11} \mathrm{~m} \end{aligned}

Question (a)

(a)

State what is meant by the solar constant.

[ 1 ]

Question (b)

(b)

Outline the physical mechanism by which some of the radiation emitted by the surface is absorbed by greenhouse gases in the atmosphere and re-radiated towards the surface.

[ 2 ]

Question (c)

(c)

Show that the average global intensity of radiation absorbed by the surface is about 240Wm−2240 \mathrm{Wm}^{-2}.

[ 2 ]

Question (d)

(d)

Determine the average intensity re-radiated by the atmosphere towards the surface. Assume that the emissivity of the surface is 0.90 .

[ 3 ]

Question 3

[Maximum number: 1]

The Moon has no atmosphere and orbits the Earth. The diagram shows the Moon with rays of light from the Sun that are incident at 90∘90^{\circ} to the axis of rotation of the Moon.

Figure for Question 3 — IB Physics SL

The albedo of the Earth's atmosphere is 0.28 . Outline why the maximum temperature of a black body on the Earth when the Sun is overhead is less than that at point A on the Moon.

All question bank results loaded