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

Question

In a simple climate model for a planet, the incoming intensity is 400 W {m}^{-2} and the radiated intensity is 300 W {m}^{-2}.

                                       

The temperature of the planet is constant. What are the reflected intensity from the planet and the albedo of the planet?

Ans:

Reflected intensity from the planet

Albedo of the planet

A

 100 W {m}^{-2}

0.25

B

100 W {m}^{-2}

0.75

C

 300 W {m}^{-2}

0.25

D

300 W {m}^{-2}

0.75

Answer/Explanation

Ans: A

Since Temperature of temperature of the planet is constant hence 

\(\Delta Q = mc\Delta T =0\)
\(\Rightarrow \) Total Incoming Intensity = Out going Intensity
Hence
\(400 = 300 + \text {reflected Intensity}\)
Hence
\(\text {reflected Intensity} = 100\; Wm^{-2}\)

\(albedo =\alpha =\frac{\text{total scattered power}}{\text{total incident power}}\)
Total scattered (reflected) Intensity \(= 100 \; W m^{-2}\)
and Total incident Intensity \(= 400 \; W m^{-2}\)
or
Total scattered (reflected) Power \(= 100 \times A \; \; W \)  A= Surface arean of planet
Total incident Power = \(400 \times A \; W\)
Hence
\(albedo =\alpha =\frac{\text{total scattered power}}{\text{total incident power}}\)
\(=\frac{100 \times A}{400 \times A} =0.25\)

Question

A black body at temperature T emits radiation with peak wavelength \lambda _p and power P. What is the temperature of the black body and the power emitted for a peak wavelength of \tfrac{\lambda _p}{2} ?

Ans:

Temperature of the black body

Power emitted by the black body

A

\frac{T}{2}

\frac{P}{16}

B

\frac{T}{2}

\frac{P}{4}

C

 2T

4P

D

 2T

16P

Answer/Explanation

Ans: D

Radiation is the transfer of energy through electromagnetic radiation.

Most energy is radiated at a specific wavelength  \(\lambda_m\) determined by Wien’s displacement law:

\(\lambda_m T= Constant = 2.9 \times 10^{-3}\)
where
\(\lambda_m =\) wavelength for which maximum intensity
\(T\) =temperature in kelvin

Hence in this given question
\(\lambda_mT =constant\)
or
\(\lambda_pT=\frac{\lambda_p}{2}T’\)
or
\(T’ = 2T\)

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
T = temperature
Hence every other parameter remain constant , as per given question
\(P = k T^4\)
hence
\(\frac{P’}{P}=\frac{T’^4}{T^4}\)
Now \(T’ = 2T\)
Hence
\(\frac{P’}{P}=\frac{(2T)^4}{T^4} =2^4=16\)
Hence \(P’ = 16\)

Question

Mars and Earth act as black bodies. The \(\frac{{{\text{power radiated by Mars}}}}{{{\text{power radiated by the Earth}}}} = p\) and \(\frac{{{\text{absolute mean temperature of the surface of Mars}}}}{{{\text{absolute mean temperature of the surface of the Earth}}}} = t\).

What is the value of \(\frac{{{\text{radius of Mars}}}}{{{\text{radius of the Earth}}}}\)?

A. \(\frac{p}{{{t^4}}}\)

B. \(\frac{{\sqrt p }}{{{t^2}}}\)

C. \(\frac{{{t^4}}}{p}\)

D. \(\frac{{{t^2}}}{{\sqrt p }}\)

Answer/Explanation

Markscheme

B

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
as per given question emissivity = 1 as planets are black bodies

\(\frac{P_{Mars}}{P_{Earth}}=\frac{\sigma \times 4\pi (R_{Mars})^2\times (T_{Mars})^4}{\sigma \times 4\pi (R_{Earth})^2\times (T_{Earth})^4}\)

\(= \frac{(R_{Mars})^2\times (T_{Mars})^4}{(R_{Earth})^2\times (T_{Earth})^4}\)

Now
\(\frac{P_{Mars}}{P_{Earth}} =p \; and\;  \frac{T_{Mars}}{T_{Earth}} =t\)
Putting these values in above equation , we get
\(\frac{(R_{Mars})^2}{(R_{Earth})^2}\times t^4=p\)
or
\(\frac{R_{Mars}}{R_{Earth}}=\sqrt{\frac{p}{t^4}}=\frac{\sqrt{p}}{t^2}\)

Question

Three gases in the atmosphere are

I. carbon dioxide (CO2)

II. dinitrogen monoxide (N2O)

III. oxygen (O2).

Which of these are considered to be greenhouse gases?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Answer/Explanation

Markscheme

A

Greenhouse gases strongly absorb infrared radiation from the atmosphere, when re-radiated in all directions, these gases account for additional warming. These gases include water vapour, carbon dioxide, methane and nitrous oxide.

Question

A black body emits radiation with its greatest intensity at a wavelength of Imax. The surface temperature of the black body doubles without any other change occurring. What is the wavelength at which the greatest intensity of radiation is emitted?

A. Imax

B. \(\frac{{{\text{I}}{{\text{}}_{{\text{max}}}}}}{{\text{2}}}\)

C. \(\frac{{{\text{I}}{{\text{}}_{{\text{max}}}}}}{{\text{4}}}\)

D. \(\frac{{{\text{I}}{{\text{}}_{{\text{max}}}}}}{{\text{16}}}\)

Answer/Explanation

Markscheme

B

\(\lambda_m T= Constant = 2.9 \times 10^{-3}\)
where
\(\lambda_m =\) wavelength for which maximum intensity
\(T\) =temperature in kelvin

Given in question

Initial condition

\(\lambda_m = I_{}max  \) , Temperature = \(T\)

2nd Condition 

\(\lambda_m =? \) , temperature = \(2T\)

Hence

\(I_{max}T=\lambda_m \times 2T\)
\(\lambda_m=\frac{I_{max}T}{2T} =\frac{I_{max}}{2}\)