Calculate the daily loss of energy by the earth, if the temperature gradient in the earth's crust is `32^(@)C` per km and mean conductivity of the rock is `0.008` of CGS unit. (Given radius of earth `=6400km`)

Options

A. `10^30 ` cal
B. `10^40 ` cal
C. `10^20 ` cal
D. `10^18 ` cal

Answer

To calculate the daily loss of energy by the Earth, we can follow these steps:

### Step 1: Understand the Given Data
- Temperature gradient in the Earth's crust: \(32 \, ^\circ C \text{ per km}\)
- Mean conductivity of the rock: \(0.008 \, \text{cal/cm/s/}^\circ C\) (CGS unit)
- Radius of the Earth: \(6400 \, \text{km}\)

### Step 2: Convert Units
1. Convert the temperature gradient from per kilometer to per centimeter:
\[ \text{Temperature Gradient} = 32 \, ^\circ C/\text{km} = \frac{32}{1000} \, ^\circ C/\text{cm} = 32 \times 10^{-5} \, ^\circ C/\text{cm} \]

2. Convert the radius of the Earth from kilometers to centimeters:
\[ R = 6400 \, \text{km} = 6400 \times 10^5 \, \text{cm} = 6.4 \times 10^7 \, \text{cm} \]

### Step 3: Calculate the Surface Area of the Earth
The surface area \(A\) of a sphere is given by the formula:
\[ A = 4\pi R^2 \]
Substituting the radius:
\[ A = 4\pi (6.4 \times 10^7)^2 = 4\pi (40.96 \times 10^{14}) \approx 4 \times 3.14 \times 40.96 \times 10^{14} \approx 5.14 \times 10^{15} \, \text{cm}^2 \]

### Step 4: Calculate the Rate of Heat Flow
The rate of heat flow (heat current) \(Q\) can be calculated using the formula:
\[ Q = k \cdot A \cdot \text{Temperature Gradient} \]
Where:
- \(k = 0.008 \, \text{cal/cm/s/}^\circ C\)
- \(\text{Temperature Gradient} = 32 \times 10^{-5} \, ^\circ C/\text{cm}\)

Substituting the values:
\[ Q = 0.008 \cdot (5.14 \times 10^{15}) \cdot (32 \times 10^{-5}) \]
Calculating this gives:
\[ Q \approx 0.008 \cdot 5.14 \cdot 32 \cdot 10^{15 - 5} = 0.008 \cdot 5.14 \cdot 32 \cdot 10^{10} \]

### Step 5: Calculate the Daily Loss of Energy
To find the total heat loss in one day, we multiply the rate of heat flow by the number of seconds in a day:
\[ \text{Total Heat Loss in One Day} = Q \cdot \text{Time} \]
Where:
- Time in seconds for one day = \(86400 \, \text{s}\)

Thus:
\[ \text{Total Heat Loss} = Q \cdot 86400 \]

### Step 6: Final Calculation
Substituting \(Q\) from above:
\[ \text{Total Heat Loss} \approx (0.008 \cdot 5.14 \cdot 32 \cdot 10^{10}) \cdot 86400 \]
Calculating this gives:
\[ \text{Total Heat Loss} \approx 1.1 \times 10^{18} \, \text{calories} \]

### Conclusion
The daily loss of energy by the Earth is approximately \(1.1 \times 10^{18} \, \text{calories}\).

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