The temperature and the surface area of the body are 227^∘ C and 0.15 m^2 respectively. If its

Name: Problem 43, Chapter 18: Arihant AIEEE Physics
Uploaded: 2021-09-24T19:14:54-08:00
Duration: 2 min 29 s
Channel: Saman Zulfiqar
Description: The temperature and the surface area of the body are 227^∘ C and 0.15 m^2 respectively. If its transmitting power is negligible and reflecting power is 0.5, then the thermal power of the body is: (Given: .σ=5.67 ×10^-8 J / m^2 s K) (a) 300 watt (b) 265.78 watt (c) 201 watt (d) 320.89 watt

step 1 In this question, it is given as the temperature and the surface area of the body is given us 2

Question

The temperature and the surface area of the body are $227^{\circ} \mathrm{C}$ and $0.15 \mathrm{~m}^{2}$ respectively. If its transmitting power is negligible and reflecting power is $0.5$, then the thermal power of the body is: (Given: $\left.\sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{m}^{2} \mathrm{~s} \mathrm{~K}\right)$ (a) 300 watt (b) $265.78$ watt (c) 201 watt (d) $320.89$ watt

   The temperature and the surface area of the body are $227^{\circ} \mathrm{C}$ and $0.15 \mathrm{~m}^{2}$ respectively. If its transmitting power is negligible and reflecting power is $0.5$, then the thermal power of the body is:
(Given: $\left.\sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{m}^{2} \mathrm{~s} \mathrm{~K}\right)$
(a) 300 watt
(b) $265.78$ watt
(c) 201 watt
(d) $320.89$ watt

Arihant AIEEE Physics

D.B. Singh 2006 Edition

Chapter 18, Problem 43

Instant Answer

Solved by Expert Saman Zulfiqar

09/24/2021

Step 1

We do this by adding 273 to the given temperature. So, $227^{\circ} \mathrm{C}$ becomes $227 + 273 = 500 \mathrm{K}$. Show more…

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The temperature and the surface area of the body are $227^{\circ} \mathrm{C}$ and $0.15 \mathrm{~m}^{2}$ respectively. If its transmitting power is negligible and reflecting power is $0.5$, then the thermal power of the body is: (Given: $\left.\sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{m}^{2} \mathrm{~s} \mathrm{~K}\right)$ (a) 300 watt (b) $265.78$ watt (c) 201 watt (d) $320.89$ watt

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Key Concepts

Temperature Conversion

In thermal radiation calculations, temperature must be expressed in Kelvin to properly apply physical laws such as the Stefan-Boltzmann law. Converting from degrees Celsius to Kelvin involves adding 273.15 to the Celsius temperature, ensuring the correct absolute temperature is used in calculations.

Emissivity

Emissivity is a measure of how efficiently a real object emits thermal radiation compared to an ideal blackbody, which has an emissivity of 1. It accounts for the fact that real surfaces emit less radiation due to factors like reflection, absorption, and transmission losses.

Stefan-Boltzmann Law

The Stefan-Boltzmann law relates the power radiated by a blackbody to its temperature and surface area, stating that the total energy radiated per unit surface area is proportional to the fourth power of its absolute temperature. It is a fundamental principle in understanding the thermal radiation from objects.

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