How long does it take for the Earth to cool down to \(3K\) 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 \(5.972\cdot10^{24}\text{ kg}\). The specific heat capacity of the Earth is \(1260 \text{ J/kg} K\). The current temperature of the Earth is \(300K\). The Earth is a perfect blackbody. The surface area of the Earth is \(5.1\cdot10^{14}m^2\). 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.

## Comments

Sort by:

TopNewestFrom the Stephan-Boltzmann Law, \[ \frac{dQ}{dt} = \sigma A T^4 = \sigma A T^4 \] \(Q \) is heat radiated by the earth, \( T \) is the temperature of the earth, \( A \) is the surface area.

By the definition of specific heat,

\[ \frac{dQ}{dT} = -mc \] \( m \) is the mass of the earth, \( c \) is the specific heat.

The negative sign is because the temperature is reducing.

Dividing,

\[ \frac{dT}{dt} = \frac{-\sigma A}{mc} \cdot T^4 \] Rearranging,

\[ \int_{300 K}^{3 K} \frac{-dT}{T^4} = \int_0^t \frac{\sigma A}{mc} dt \]

\[ \frac{1}{3} \left(\frac{1}{T^3}\right)\biggl|_{300}^{3} = \frac{\sigma A t}{mc} \] so,

\[ t = \frac{mc}{3\sigma A} \left(\frac{1}{27} - \frac{1}{27 \cdot 10^6}\right) \]

\[ t \approx 3.23 \cdot 10^{18} \text{ seconds } \approx 10^{11} \text{ years } \] That is a long time! – Ameya Daigavane · 1 year, 2 months ago

Log in to reply

– Rishabh Tiwari · 1 year, 2 months ago

Amazing solution sir , +1!Log in to reply

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 · 1 year, 2 months ago

Log in to reply