“2019年IOAA理论第5题-CMB烤箱”的版本间的差异

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(创建页面,内容为“==英文题目== 5. Cosmic Microwave Background Oven (10 p) Since the human body is made mostly of water, it is very efficient at absorbing microwave photons. Ass…”)
 
 
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{{需要解答}}
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==英文题目==
 
==英文题目==
  
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Assume that an astronaut’s body is a perfect spherical absorber with mass of $$𝑚 = 60 kg$$, and its
 
Assume that an astronaut’s body is a perfect spherical absorber with mass of $$𝑚 = 60 kg$$, and its
average density and heat capacity are the same as for pure water, i.e. $$𝜌 = 1000 kg m^−3$$ and $$𝐶 =
+
average density and heat capacity are the same as for pure water, i.e. $$𝜌 = 1000 kg \ m^{−3}$$ and $$𝐶 =
4200 J kg^−1 K^−1$$.
+
4200 J kg^{−1} K^{−1}$$.
  
 
a) What is the approximate rate, in watts, at which an astronaut in intergalactic space would absorb
 
a) What is the approximate rate, in watts, at which an astronaut in intergalactic space would absorb
 
radiative energy from the Cosmic Microwave Background (CMB)? The spectral energy
 
radiative energy from the Cosmic Microwave Background (CMB)? The spectral energy
distribution of CMB can be approximated by blackbody radiation of temperature $$𝑇_CMB =
+
distribution of CMB can be approximated by blackbody radiation of temperature $$𝑇_{CMB} =
 
2.728 K$$. (5 p)
 
2.728 K$$. (5 p)
  
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c) Ignoring other energy inputs and outputs, how long would it take for the CMB to raise the
 
c) Ignoring other energy inputs and outputs, how long would it take for the CMB to raise the
 
astronaut’s temperature by $$Δ𝑇 = 1 K$$? (2 p)
 
astronaut’s temperature by $$Δ𝑇 = 1 K$$? (2 p)
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==中文题目==
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5.宇宙微波背景烤箱(10 p)
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由于人体主要由水构成,因此在吸收微波光子方面非常高效。
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 +
假设宇航员的身体是一个完美的球形吸收体,质量$$𝑚= 60 kg$$,其平均密度和热容量与纯水相同,即$$𝜌 = 1000 kg \ m^{−3}$$ 和 $$𝐶 =
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4200 J kg^{−1} K^{−1}$$.
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a)星系际空间中的宇航员从宇宙微波背景(CMB)吸收辐射能量的大致速率(以瓦为单位)是多少? CMB的光谱能量分布可通过温度为$$𝑇_{CMB}= 2.728 K$$的黑体辐射来近似。(5 p)
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b)宇航员每秒大约吸收多少个CMB光子? (3 p)
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c)忽略其他能量输入和输出,CMB将宇航员的温度提高$$Δ𝑇= 1 K$$需要多长时间? (2 p)
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<br />
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== 解答 ==
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a)CMB的辐射功率满足斯特藩—玻尔兹曼定律
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$$P=S\sigma T^4$$
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其中P表示接收到辐射的功率,σ表示斯特藩—玻尔兹曼常量,S为接收面积,T为辐射源温度。
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根据人体质量密度,可求出这位球形老哥的半径为
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R=0.24m
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因此其表面积为
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$$S=4πR^2 m^2$$
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代入数据计算
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可得$$P=3.1x10^-6W$$
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<br />

2022年4月10日 (日) 15:36的最新版本

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英文题目

5. Cosmic Microwave Background Oven (10 p)

Since the human body is made mostly of water, it is very efficient at absorbing microwave photons.

Assume that an astronaut’s body is a perfect spherical absorber with mass of $$𝑚 = 60 kg$$, and its average density and heat capacity are the same as for pure water, i.e. $$𝜌 = 1000 kg \ m^{−3}$$ and $$𝐶 = 4200 J kg^{−1} K^{−1}$$.

a) What is the approximate rate, in watts, at which an astronaut in intergalactic space would absorb radiative energy from the Cosmic Microwave Background (CMB)? The spectral energy distribution of CMB can be approximated by blackbody radiation of temperature $$𝑇_{CMB} = 2.728 K$$. (5 p)

b) Approximately how many CMB photons per second would the astronaut absorb? (3 p)

c) Ignoring other energy inputs and outputs, how long would it take for the CMB to raise the astronaut’s temperature by $$Δ𝑇 = 1 K$$? (2 p)

中文题目

5.宇宙微波背景烤箱(10 p)

由于人体主要由水构成,因此在吸收微波光子方面非常高效。

假设宇航员的身体是一个完美的球形吸收体,质量$$𝑚= 60 kg$$,其平均密度和热容量与纯水相同,即$$𝜌 = 1000 kg \ m^{−3}$$ 和 $$𝐶 = 4200 J kg^{−1} K^{−1}$$.

a)星系际空间中的宇航员从宇宙微波背景(CMB)吸收辐射能量的大致速率(以瓦为单位)是多少? CMB的光谱能量分布可通过温度为$$𝑇_{CMB}= 2.728 K$$的黑体辐射来近似。(5 p)

b)宇航员每秒大约吸收多少个CMB光子? (3 p)

c)忽略其他能量输入和输出,CMB将宇航员的温度提高$$Δ𝑇= 1 K$$需要多长时间? (2 p)


解答

a)CMB的辐射功率满足斯特藩—玻尔兹曼定律

$$P=S\sigma T^4$$

其中P表示接收到辐射的功率,σ表示斯特藩—玻尔兹曼常量,S为接收面积,T为辐射源温度。

根据人体质量密度,可求出这位球形老哥的半径为

R=0.24m

因此其表面积为

$$S=4πR^2 m^2$$

代入数据计算

可得$$P=3.1x10^-6W$$