Heat Transfer - Radiation: Practice Problem & Solution
The sun delivers about \( 1400 \text{ W/m}^2 \) of electromagnetic flux to Earth's surface. If the flux falls on a roof of dimensions \( 8\text{ m} \times 20\text{ m} \) normally, then total power incident on the roof will be
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Heat Transfer - Radiation. Understanding the underlying formula is key to arriving at the correct answer below:
Power is given by the product of intensity and area: \( P = I \times A \). Here \( I = 1400 \text{ W/m}^2 \) and \( A = 8\text{ m} \times 20\text{ m} = 160 \text{ m}^2 \). Therefore, \( P = 1400 \times 160 = 224,000 \text{ W} = 224 \text{ kW} \).
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