Cooling of the Earth

How long does it take for the Earth to cool down to 3K3K if the sun was gone?

Assumptions: The Earth only loses heat to space and does not gain any heat. Heat transfer occurs only through radiation. The mass of the Earth is 5.9721024 kg5.972\cdot10^{24}\text{ kg}. The specific heat capacity of the Earth is 1260 J/kgK1260 \text{ J/kg} K. The current temperature of the Earth is 300K300K. The Earth is a perfect blackbody. The surface area of the Earth is 5.11014m25.1\cdot10^{14}m^2. Do not take the latent heat of fusion or vaporisation into account. Assume heat is always evenly distributed around the Earth and the Earth is a uniform sphere made of the same material with the specific heat capacity as stated above.

Note by A Brilliant Member
3 years, 2 months ago

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From the Stephan-Boltzmann Law, dQdt=σAT4=σAT4 \frac{dQ}{dt} = \sigma A T^4 = \sigma A T^4 QQ is heat radiated by the earth, T T is the temperature of the earth, A A is the surface area.
By the definition of specific heat,
dQdT=mc \frac{dQ}{dT} = -mc m m is the mass of the earth, c c is the specific heat.
The negative sign is because the temperature is reducing.
Dividing,
dTdt=σAmcT4 \frac{dT}{dt} = \frac{-\sigma A}{mc} \cdot T^4 Rearranging,
300K3KdTT4=0tσAmcdt \int_{300 K}^{3 K} \frac{-dT}{T^4} = \int_0^t \frac{\sigma A}{mc} dt

13(1T3)3003=σAtmc \frac{1}{3} \left(\frac{1}{T^3}\right)\biggl|_{300}^{3} = \frac{\sigma A t}{mc} so,
t=mc3σA(127127106) t = \frac{mc}{3\sigma A} \left(\frac{1}{27} - \frac{1}{27 \cdot 10^6}\right)

t3.231018 seconds 1011 years  t \approx 3.23 \cdot 10^{18} \text{ seconds } \approx 10^{11} \text{ years } That is a long time!

Ameya Daigavane - 3 years, 2 months ago

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Amazing solution sir , +1!

Rishabh Tiwari - 3 years, 2 months ago

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Hey Jerry , its interesting to know that we can even calculate this , Can you explain how ? Anyone ? @Pranshu Gaba , @Swapnil Das , @Arjen Vreugdenhil please help .

Rishabh Tiwari - 3 years, 2 months ago

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