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string | index
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string | answer_type
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(1) Find the expression of the solar constant $S_0$.
(2) Calculate the value of $S_0$ (expressed in $W/m^2$).
|
IPhO_2024
|
[0.4, 0.2]
|
text-only
|
[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[Earth as a Blackbody]
In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.
|
0
|
[["Award 0.4 pt if the answer gives the correct expression for the solar constant: $S_0 = \\sigma T_S^4 (\\frac{R_S}{d})^2$. Partial points: award 0.1 pt if the answer gives the incorrect expression but realizes energy conservation. Otherwise, award 0 pt.", "Award 0.2 pt if the answer gives the correct numerical value of the solar constant: $1.35 \\times 10^{3} \\frac{W}{m^2}$. Partial points: award 0.1 pt if the numerical answer falls within the acceptable error range of the correct value but contains more than four significant figures. Otherwise, award 0 pt."]]
|
["Expression", "Numerical Value"]
|
[null, "$W/m^2$"]
|
Thermodynamics
|
None.
|
["\\boxed{$S_0 = \\sigma T_S^4 (\\frac{R_S}{d})^2$}", "\\boxed{$1.35 \\times 10^{3}$}"]
|
IPhO_2024_1_A_1
| Not supported with pagination yet |
|
(1) Find the expression of the Earth's temperature $T_{\mathrm{E}}$.
(2) Calculate the value of $T_{\mathrm{E}}$ (expressed in $\mathrm{K}$).
|
IPhO_2024
|
[0.4, 0.2]
|
text-only
|
[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[Earth as a Blackbody]
In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.
|
1
|
[["Award 0.4 pt if the answer gives the correct expression for the Earth's temperature: $T_{\\mathrm{E}} = (\\frac{S_0}{4 \\sigma})^{1/4} = \\sqrt{\\frac{R_S}{2d}} T_S$. Partial points: award 0.1 pt if the answer gives the incorrect expression but realizes energy balance. Otherwise, award 0 pt.", "Award 0.2 pt if the answer gives the correct numerical value of the Earth's temperature: $278 \\mathrm{K}$. Partial points: award 0.1 pt if the numerical answer falls within the acceptable error range of the correct value but contains more than four significant figures. Otherwise, award 0 pt."]]
|
["Expression", "Numerical Value"]
|
[null, "$\\mathrm{K}$"]
|
Thermodynamics
|
None.
|
["\\boxed{$T_{\\mathrm{E}} = (\\frac{S_0}{4 \\sigma})^{1/4} = \\sqrt{\\frac{R_S}{2d}} T_S$}", "\\boxed{278}"]
|
IPhO_2024_1_A_2
| Not supported with pagination yet |
|
Find the function $f(x)$.
|
IPhO_2024
|
[0.4]
|
text-only
|
[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[Earth as a Blackbody]
In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.
|
2
|
[["Award 0.4 pt if the answer gives the correct expression for the function $f(x)$: $f(x) = 5(1-e^{-x})-x$ (Equivalent forms are also correct, e.g., $f(x) = (5-x)e^x-5 = 5e^x -5 - x e^x$). Otherwise, award 0 pt."]]
|
["Expression"]
|
[null]
|
Thermodynamics
|
None.
|
["\\boxed{$f(x) = 5(1-e^{-x})-x$}"]
|
IPhO_2024_1_A_3
| Not supported with pagination yet |
|
(1) Calculate the numerical value of $x_{\mathrm{m}}$.
(2) From this value $x_{\mathrm{m}}$, find the value of $b$ (expressed in $\mathrm{nm} \cdot K$).
|
IPhO_2024
|
[0.3, 0.1]
|
text-only
|
[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[Earth as a Blackbody]
In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.
|
3
|
[["Award 0.3 pt if the answer gives the correct numerical value of $x_{\\mathrm{m}}$ within the range of $[4.96, 4.97]$. Partial points: award 0.2 pt if the answer gives a value of $x_{\\mathrm{m}}$ within the range of $[4.96, 4.97]$ but contains more than four significant figures. Otherwise, award 0 pt.", "Award 0.1 pt if the answer gives the correct numerical value of $b$ within the range of $[2.89 \\times 10^{6}, 2.90 \\times 10^{6}]$ $\\mathrm{nm} \\cdot K$. Otherwise, award 0 pt."]]
|
["Numerical Value", "Numerical Value"]
|
[null, "$\\mathrm{nm} \\cdot K$"]
|
Thermodynamics
|
None.
|
["\\boxed{$[4.96, 4.97]$}", "\\boxed{$[2.89 \\times 10^{6}, 2.90 \\times 10^{6}]$}"]
|
IPhO_2024_1_A_4
| Not supported with pagination yet |
|
(1) Find $\lambda_{\text{max}}^{\text{Sun}}$ for the Sun (expressed in $\mathrm{nm}$).
(2) Find $\lambda_{\text{max}}^{\text{Earth}}$ for the Earth (expressed in $\mathrm{nm}$).
|
IPhO_2024
|
[0.1, 0.1]
|
text-only
|
[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[Earth as a Blackbody]
In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.
|
4
|
[["Award 0.1 pt if the answer gives the correct numerical value of $\\lambda_{\\text{max}}^{\\text{Sun}}$ within the range of $[501, 502] \\mathrm{nm}$ (Equivalent form of $[5.01 \\times 10^{2} \\mathrm{nm}, 5.02 \\times 10^{2} \\mathrm{nm}]$ is also correct). Otherwise, award 0 pt.", "Award 0.1 pt if the answer gives the correct numerical value of $\\lambda_{\\text{max}}^{\\text{Earth}}$ as $1.04 \\times 10^{4} \\mathrm{nm}$ (Equivalent forms are also correct). Otherwise, award 0 pt."]]
|
["Numerical Value", "Numerical Value"]
|
["$\\mathrm{nm}$", "$\\mathrm{nm}$"]
|
Thermodynamics
|
None.
|
["\\boxed{$[5.01 \\times 10^{2}, 5.02 \\times 10^{2}]$}", "\\boxed{$1.04 \\times 10^{4}$}"]
|
IPhO_2024_1_A_5
| Not supported with pagination yet |
|
(1) Find the expression of $\gamma$.
(2) Determine the value of $\gamma$.
|
IPhO_2024
|
[0.6, 0.2]
|
text+data figure
|
[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[Earth as a Blackbody]
In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.
As shown in the Figure 1, the functions $\gamma \tilde{u}_{\mathrm{S}}(\lambda)$ and $u\left(\lambda, T_{\mathrm{E}}\right)$ are plotted versus $\lambda$, where $\gamma$ is a dimensionless coefficient to rescale $\tilde{u}_{S}(\lambda)$ such that the values of the two peaks coincide.
[figure1]
Figure 1. The plot of $u(\lambda, T_{\mathrm{E}})$ (red) and $\gamma \tilde{u}_{S}(\lambda)$ (blue) versus $\lambda$.
|
5
|
[["Award 0.6 pt if the answer gives the correct expression for $\\gamma$: $\\gamma = (\\frac{d}{R_S})^2 \\times (\\frac{T_E}{T_S})^5 = (\\frac{d}{R_S})^2 \\times (\\frac{\\lambda_S}{\\lambda_E})^5$. Partial points: award 0.3 pt if the answer realizes that $\\tilde{u}_S = (\\frac{R_S}{d})^2 u_S(\\lambda)$. Otherwise, award 0 pt.", "Award 0.2 pt if the answer gives the correct numerical value of $\\gamma$ within the range of $[1.20 \\times 10^{-2}, 1.21 \\times 10^{-2}]$. Partial points: award 0.1 pt if the numerical answer falls within the acceptable error range of the correct value but contains more than four significant figures. Otherwise, award 0 pt."]]
|
["Expression", "Numerical Value"]
|
[null, null]
|
Thermodynamics
|
None.
|
["\\boxed{$\\gamma = (\\frac{d}{R_S})^2 \\times (\\frac{T_E}{T_S})^5 = (\\frac{d}{R_S})^2 \\times (\\frac{\\lambda_S}{\\lambda_E})^5$}", "\\boxed{$[1.20 \\times 10^{-2}, 1.21 \\times 10^{-2}]$}"]
|
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IPhO_2024_1_A_6
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Assume that $\varepsilon=1$ and $r_{\mathrm{E}}=0$. (1) Find the expression of the Earth's temperature $T_{\mathrm{E}}$; (2) Find the expression of the atmosphere's temperature $T_{\mathrm{A}}$; (3) Calculate the numerical value of $T_{\mathrm{E}}$ (expressed in $K$); (4) (3) Calculate the numerical value of $T_{\mathrm{A}}$ (expressed in $K$).
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IPhO_2024
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[0.4, 0.4, 0.1, 0.1]
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text+illustration figure
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[The Greenhouse Effect]
In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.
All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T) d \lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda+d \lambda$. According to Planck's theory of blackbody radiation, we have
$$u(\lambda, T)=\frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{\mathrm{B}} T}\right)-1}$$ (Equation 1), in which $h c=1.24 \times 10^{3} \mathrm{eV} \cdot \mathrm{nm}$ and $k_{\mathrm{B}}=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}$. The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T=b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b=\frac{h c}{x_{\mathrm{m}} k_{\mathrm{B}}}$, where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x)=0$; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T)=\sigma T^{4}$ where $\sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}$. Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.
Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}}=5.77 \times 10^{3} \mathrm{K}$. The Sun's radius is $R_{\mathrm{S}}=6.96 \times 10^{8} \mathrm{m}$ and the average distance between the Earth and the Sun is $d=1.50 \times 10^{11} \mathrm{m}$. We denote by $\tilde{u}_{S}(\lambda)$, the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_{0}=\int \tilde{u}_{S}(\lambda) d \lambda$, is called the solar constant.
In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.
[The Greenhouse Effect]
In this part, we introduce a simple model in which the Earth's atmosphere is modeled as a thin layer at a small distance above the Earth's surface so that the difference between the area of the atmosphere's layer and the area of the Earth's surface can be neglected (see Figure 1). In what follows assume that the major part of the thermal radiation from the Earth and the Sun are emitted at wavelengths near the $\lambda_{\max}$ for each one. Also assume that the "atmosphere layer" reflects a fraction $r_{\mathrm{A}}=0.255$ of the visible-ultraviolet radiation incident from above or below, and completely transmits the rest. Assume that the atmosphere does not reflect any part of the infrared radiation, however, it absorbs a fraction $\varepsilon$ of the infrared radiation and transmits the rest. This behavior, known as the greenhouse effect, changes the average temperature of the Earth. The Earth's surface, on the other hand, reflects a fraction $r_{E}$ of the visible-ultraviolet radiation and absorbs the rest of this radiation and all the infrared radiation.
[figure1]
Figure 1. Thermal flows between the Earth and the atmosphere.
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6
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[["Award 0.8 pt if the answer gives two correct expressions: $T_{\\mathrm{E}} = \\left( \\frac{(1-r_A) \\frac{S_0}{2}}{\\sigma} \\right)^{1/4}$, and $T_{\\mathrm{A}} = \\left( \\frac{(1-r_A) \\frac{S_0}{4}}{\\sigma} \\right)^{1/4}$. Partial points: award 0.6 pt if the answer gives only one of the two expressions correctly; or award 0.2 pt if the answer gives no correct expressions but realizes each energy balance relation. Otherwise, award 0 pt.", "Award 0.2 pt if the answer gives the correct numerical value of $T_{\\mathrm{E}}$ as $3.07 \\times 10^{2} K$, and $T_{\\mathrm{A}}$ as $2.58 \\times 10^{2} K$. Partial points: award 0.1 pt if the numerical answer falls within the acceptable error range of the correct value but contains more than four significant figures. Otherwise, award 0 pt."]]
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["Expression", "Expression", "Numerical Value", "Numerical Value"]
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[null, null, "K", "K"]
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Thermodynamics
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None.
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["\\boxed{$T_{\\mathrm{E}} = \\left( \\frac{(1-r_A) \\frac{S_0}{2}}{\\sigma} \\right)^{1/4}$}", "\\boxed{$T_{\\mathrm{A}} = \\left( \\frac{(1-r_A) \\frac{S_0}{4}}{\\sigma} \\right)^{1/4}$}", "\\boxed{$T_E = 3.07 \\times 10^2$}", "\\boxed{$T_A = 2.58 \\times 10^2$}"]
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
|
IPhO_2024_1_B_1
| |
"(1) Determine the albedo, $\\alpha$, in terms of $r_{\\mathrm{E}}$ and $r_{\\mathrm{A}}$. \n(2) Cal(...TRUNCATED) |
IPhO_2024
|
[1.4, 0.2]
|
text+illustration figure
| "[The Greenhouse Effect] \n\nIn 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the N(...TRUNCATED) |
7
| "[[\"Award 1.4 pt if the answer gives the correct expression for $\\\\alpha$: $\\\\alpha = r_{\\\\ma(...TRUNCATED) |
["Expression", "Numerical Value"]
|
[null, null]
|
Thermodynamics
|
None.
| "[\"\\\\boxed{$\\\\alpha = r_{\\\\mathrm{A}} + \\\\frac{(1-r_{\\\\mathrm{A}})^{2} r_{\\\\mathrm{E}}}(...TRUNCATED) | "/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDA(...TRUNCATED) |
IPhO_2024_1_B_2
| |
"(1) Express the Earth's temperature $T_{\\mathrm{E}}$ in terms of $\\sigma, \\alpha, S_{0}$, and $\(...TRUNCATED) |
IPhO_2024
|
[0.6, 0.4]
|
text+illustration figure
| "[The Greenhouse Effect] \n\nIn 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the N(...TRUNCATED) |
8
| "[[\"Award 0.6 pt if the answer gives the correct expression for $T_{\\\\mathrm{E}}$: $T_{\\\\mathrm(...TRUNCATED) |
["Expression", "Numerical Value"]
|
[null, null]
|
Thermodynamics
|
None.
| "[\"\\\\boxed{$T_{\\\\mathrm{E}} = \\\\left[\\\\frac{(1-\\\\alpha)}{2 \\\\sigma(2-\\\\epsilon)} S_{0(...TRUNCATED) | "/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDA(...TRUNCATED) |
IPhO_2024_1_B_3
| |
"(1) Find the expression of $\\frac{\\mathrm{d} T_{\\mathrm{E}}}{\\mathrm{d} \\varepsilon}$. \n(2) D(...TRUNCATED) |
IPhO_2024
|
[0.6, 0.2]
|
text+illustration figure
| "[The Greenhouse Effect] \n\nIn 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the N(...TRUNCATED) |
9
| "[[\"Award 0.6 pt if the answer gives the correct expression for $\\\\frac{\\\\mathrm{d} T_{\\\\math(...TRUNCATED) |
["Expression", "Numerical Value"]
|
[null, "K"]
|
Thermodynamics
|
None.
| "[\"\\\\boxed{$\\\\frac{\\\\mathrm{d} T_{\\\\mathrm{E}}}{\\\\mathrm{d} \\\\varepsilon} = \\\\frac{1}(...TRUNCATED) | "/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDA(...TRUNCATED) |
IPhO_2024_1_B_4
|
End of preview. Expand
in Data Studio
🏆 New (Sep. 16): We launched "PhyArena", a physics reasoning leaderboard incorporating the HiPhO benchmark.
🌐 Introduction
HiPhO (High School Physics Olympiad Benchmark) is the first benchmark specifically designed to evaluate the physical reasoning abilities of (M)LLMs on real-world Physics Olympiads from 2024–2025.
✨ Key Features
- Up-to-date Coverage: Includes 13 Olympiad exam papers from 2024–2025 across international and regional competitions.
- Mixed-modal Content: Supports four modality types, spanning from text-only to diagram-based problems.
- Professional Evaluation: Uses official marking schemes for answer-level and step-level grading.
- Human-level Comparison: Maps model scores to medal levels (Gold/Silver/Bronze) and compares with human performance.
🏆 IPhO 2025 (Theory) Results
- Top-1 Human Score: 29.2 / 30.0
- Top-1 Model Score: 22.7 / 29.4 (Gemini-2.5-Pro)
- Gold Threshold: 19.7
- Silver Threshold: 12.1
- Bronze Threshold: 7.2
Although models like Gemini-2.5-Pro and GPT-5 achieved gold-level scores, they still fall noticeably short of the very best human contestants.
📊 Dataset Overview
HiPhO contains:
- 13 Physics Olympiads
- 360 Problems
- Categorized across:
- 5 Physics Fields: Mechanics, Electromagnetism, Thermodynamics, Optics, Modern Physics
- 4 Modality Types: Text-Only, Text+Illustration Figure, Text+Variable Figure, Text+Data Figure
- 6 Answer Types: Expression, Numerical Value, Multiple Choice, Equation, Open-Ended, Inequality
Evaluation is conducted using:
- Answer-level and step-level scoring, aligned with official marking schemes
- Exam score as the evaluation metric
- Medal-based comparison, using official thresholds for gold, silver, and bronze
🖼️ Modality Categorization
- 📝 Text-Only (TO): Pure text, no figures
- 🎯 Text+Illustration Figure (TI): Figures illustrate physical setups
- 📐 Text+Variable Figure (TV): Figures define key variables or geometry
- 📊 Text+Data Figure (TD): Figures show plots, data, or functions absent from text
As models move from TO → TD, performance drops sharply—highlighting the challenges of visual reasoning.
📈 Main Results
- Closed-source reasoning MLLMs lead the benchmark, earning 6–12 gold medals (Top 5: Gemini-2.5-Pro, Gemini-2.5-Flash-Thinking, GPT-5, o3, Grok-4)
- Open-source MLLMs mostly score at or below the bronze level
- Open-source LLMs demonstrate stronger reasoning and generally outperform open-source MLLMs
📥 Download
- Dataset & Annotations: https://huggingface.co/datasets/SciYu/HiPhO
- GitHub Repository: https://github.com/SciYu/HiPhO
- 📄 Paper: https://arxiv.org/abs/2509.07894
- 📧 Contact: fangchenyu@link.cuhk.edu.cn
🔖 Citation
@article{hipho2025,
title={HiPhO: How Far Are (M)LLMs from Humans in the Latest High School Physics Olympiad Benchmark?},
author={Yu, Fangchen and Wan, Haiyuan and Cheng, Qianjia and Zhang, Yuchen and Chen, Jiacheng and Han, Fujun and Wu, Yulun and Yao, Junchi and Hu, Ruilizhen and Ding, Ning and Cheng, Yu and Chen, Tao and Bai, Lei and Zhou, Dongzhan and Luo, Yun and Cui, Ganqu and Ye, Peng},
journal={arXiv preprint arXiv:2509.07894},
year={2025}
}
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