# IBDP Physics 8.2 – Thermal energy transfer: IB Style Question Bank HL Paper 1

### Question

The dashed line on the graph shows the variation with wavelength of the intensity of solar radiation before passing through the Earth’s atmosphere.

The solid line on the graph shows the variation with wavelength of the intensity of solar radiation after it has passed through the Earth’s atmosphere.

Which feature of the graph helps explain the greenhouse effect?

A. Infrared radiation is absorbed at specific wavelengths.

B. There is little absorption at infrared wavelengths.

C. There is substantial absorption at visible wavelengths.

D. There is little absorption at UV wavelengths.

### Markscheme

A

The Atmosphere And Air Mass

The atmosphere scatters and absorbs some of the Sun’s energy that is incident on the Earth’s surface. Scattering of radiation by gaseous molecules (e.g. O2, O3, H2O and CO2), that are a lot smaller than the wavelengths of the radiation, is called Rayleigh scattering. Roughly half of the radiation that is scattered is lost to outer space, the remaining half is directed towards the Earth’s surface from all directions as diffuse radiation. Because of absorption by oxygen and ozone molecules the shortest wavelength that reaches the Earth’s surface is approximately 0.29 µm. Other gas molecules absorbed difference wavelengths as indicated in figure below.

### Question

The average albedo of glacier ice is 0.25.

What is $$\frac{{{\text{power absorbed by glacier ice}}}}{{{\text{power reflected by glacier ice}}}}$$?

A.  0.25

B.  0.33

C.  2.5

D.  3.0

### Markscheme

D

$$albedo =\alpha =\frac{\text{total reflected power}}{\text{total incident power}}$$

Here you can see total incident power = total reflected power + Total  absorbed power

Incident Power = Reflected + absorbed
Now
$$albedo =\alpha =\frac{\text{total reflected power}}{\text{total incident power}}$$
$$0.25 = \frac{\text{total reflected power}}{\text{total incident power}}$$
$$0.25 =\frac{reflected}{reflected + absorbed}$$  Since incident =reflected + absorbed
$$0.25 \times (reflected + absorbed) =reflected$$
$$0.25 \times (absorbed) =0.75 times reflected$$
$$\frac{absorbed}{reflected} =\frac{0.75}{0.25}=3$$

### Question

An object can lose energy through

I.    conduction
II.   convection

What are the principal means for losing energy for a hot rock resting on the surface of the Moon?

A.  I and II only

B.  I and III only

C.  II and III only

D.  I, II and III

## Markscheme

B

### Question

X and Y are two spherical black-body radiators that emit the same total power. The absolute temperature of X is half that of Y.

What is $$\frac{{{\text{radius of X}}}}{{{\text{radius of Y}}}}$$?

A. 4

B. 8

C. 16

D. 32

### Markscheme

A

Black-body radiation The power radiated by a body is governed by Stefan-Boltzman law
$$P=\varepsilon \sigma AT^4$$
Where
$$P= power$$
$$\varepsilon =emissivity = 1$$ for Black body
$$\sigma =$$ Stefan-Boltzmann constant
$$A= Area = 4\pi r^2$$
T = temperature

$$\frac{P_X}{P_Y} =\frac{\varepsilon \sigma A_X T_X1^4}{\varepsilon \sigma A_Y T_Y^4}$$
$$=\frac{\varepsilon \sigma 4\pi R_X^2 T_X^4}{\varepsilon \sigma 4\pi R_Y^2T_Y^4}$$
Given $$X$$  and $$Y$$  to be black body and $$P_X =P_Y$$  and  $$T_X = \frac{T_Y}{2}$$
hence $$\varepsilon =1$$
Putting these values in above equation , we get

$$\frac{ R_X^2 T_X^4}{ R_Y^2T_Y^4} =1$$
or
$$\frac{R_X^2}{R_Y^2} =\frac{T_Y^4}{T_X^4}=2^4$$
or
$$\frac{R_X}{R_Y}=2^2=4$$

### Question

The solar constant is the intensity of the Sun’s radiation at

A. the surface of the Earth.
B. the mean distance from the Sun of the Earth’s orbit around the Sun.
C. the surface of the Sun.
D. 10km above the surface of the Earth.

## Markscheme

B

#### SOLAR CONSTANT

Solar constant is the solar radiation incident normally per second on one square metre at the mean distance of the earth from the sun in free space.
It is given by S = 1.34 × 109 Jm–2s–1.
Temperature of sun is given T4 = S/σ(R/r)2, where R is mean distance of the earth from the sun and r is the radius of the sun.
we know the temperature of the Sun we can use the Stefan–Boltzmann law to calculate the power radiated per unit area:

$$P= A\sigma T^4$$
$$\frac{P}{A} = 5.67\times 10^{-8} \times (5800)^4$$
$$=6.42\times 10^7$$  Watt
Radius of  $$Sun = 6.9 \times 10^8 \; m$$
Hence $$A = 4\pi r^2 = 6 \times 10^{18} \;m^2$$
Hence $$P = 3.9 \times 10^{26}\;W$$
Now distance of Earth from $$Sun = 1.5 \times 10^{11} \; m$$
so the power per unit area at the Earth will be:
$$I = \frac{P}{4\pi r^2}=\frac{3.9 \times 10^{26} }{4\pi(1.5 \times 10^{11})^2}$$
$$=1400 \; W\;m^{-2}$$