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 Former Brilliant Member
5 years ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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.
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 - 5 years ago

Log in to reply

Amazing solution sir , +1!

Rishabh Tiwari - 5 years ago

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 - 5 years ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...