The Earth cools by radiating energy into space (at ~0 K). While it was once an e

Question

The Earth cools by radiating energy into space (at ~0 K). While it was once an entirely molten ball, luckily it is now quite habitable.

a.) Using the following data, estimate the age of the Earth (in solid phase).

In this crude approximation, you should:
- estimate that the Earth has a uniform temperature throughout at any given time.
- NOT use the Newton's cooling approximation (exponential)-- think about a better limit to use.

The Earth cools by radiating energy into space (at ~0 K). While it was once an entirely molten ball, luckily it is now quite habitable. Using the following data, estimate the age of the Earth (in solid phase). In this crude approximation, you should: - estimate that the Earth has a uniform temperature throughout at any given time - NOT use the Newton's cooling approximation (exponential)-- think about a better limit to use. Physical data: Mass of Earth = 5.97·1024 kg Radius of Earth = 6.38·106 m Approximate specific heat of Earth (average) = 728 J/(kg·K) Approximate minimum melting temperature of Earth (silicate rock) = 1687 K Current average crust temperature (seen in many1 caves) = 15 oC emissivity of the Earth's surface (varies by surface conditions) = 0.9 Estimated age of the Earth (as a solid) =

Explanation / Answer

we have formula

the rate of cooling of earth E1 = (M * Cp * delT)/t

here "t" is time is taken as the age of earth

and

let T1= crust tempatarure =15 C

T1 = 273+15= 288 K

and T2=melting temparature of the earth =1687 K

delT= T2-T1=1687- 288=1399 k

M is the mass of earth.

Cp is specific heat of earth.

so,

E1=(5.97*10^24*728*1399)/t

E1=(6080277.84*10^24)/t joule/m^2/sec...............................(1)

now

from the stefan's-Boltzman law,

rate of energy radiated from earth is

E2=e*sigma*A*(T_earthsurface^4 - T_space^4)

here,

the temperature of earth surface varies from 1687 C to 15 C with respect time

hence we take auerage temparature,

T_earth surface = (1687 +15)/2 = 851 K

T_space = 0 K

e is emissivity = 0.9

stefan constant sigma =5.67*10^-8 Watt*m^2/K^4

area A=4*pi*R^2

where R is the radius of the earth

so,

E2=(0.9)*(5.67*10^-8)*4*3.14*(6.38*10^6)^2 *(851^4 - 0)

E2=1374722194362345.3007*10^4 joule/m^2/sec .........................(2)

from (1) and (2)

the rate of coolinh of earth is equals to the rate of energy radiated from the earth

hence,

(6080277.84*10^24)/t=1374722194362345.3007*10^4

t=4.422914*10^11 sec

t=1.402496783*10^4 years (approximately)

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