Heat Transfer Numerical on Stefan’s Law and Wien’s Law Class-11 Nootan ISC Physics Solutions Ch-18. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Heat Transfer Numerical on Stefan’s Law and Wien’s Law Class-11 Nootan ISC Physics Solutions
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-18 | Heat Transfer |
| Topics | Numerical on Stefan’s Law and Wien’s Law |
| Academic Session | 2024-2025 |
Numerical on Stefan’s Law and Wien’s Law
( Heat Transfer Class-11 Nootan ISC Physics Solutions Ch-17 of Kumar and Mittal Physics )
Que-18: The rate of radiation from a black body at 0°C is Q J s^-1. Find the rate of radiation by the same black body at 273°C.
Ans: E1 / E2 = (T1 / T2)^4
=> Q / E2 = (273 + 0 / 273 + 273)² = 1 / 16
=> E2 = 16 Q J s^-1
Que-19: A black body radiates 5000 joules of energy per second at 227°C temperature. What will be the rate of radiation of energy at 727°C temperature?
Ans: E1 / E2 = (T1 / T2)^4
=> 5000 / E2 = (273 + 227 / 273 + 727)^4 = (500 / 1000)^4 = 1/16
=> E2 = 80000 J s^-1
Que-20: A black body at 127°C temperature is radiating energy from its surface at a rate of 1.0 x 10^6 joule per second per metre². Find that temperature of the black body at which the rate of energy radiation will be 16.0 x 10^6 joule per second per metre².
Ans: E1 / E2 = (T1 / T2)^4
=> 1 / 16 = (273 + 127 / 273 + T2)^4
=> 1 / 2 = 400 / 273 + T2
=> T2 = 800 – 273 = 527 °C
Que-21: A black body radiates 1 kilojoule (kJ) energy per second at a temperature of 27°C. Find the temperature at which it will radiate 16 kilojoule energy per second.
Ans: E1 / E2 = (T1 / T2)^4
=> 1 / 16 = (273 + 27 / 273 + T2)^4 = (300 / 273 + T2)^4
=> 300 / 273 + T2 = 1 / 2
=> T2 = 600 – 273 = 327 °C
Que-22: The luminosity of Rigel star in Orion constellation is 17,000 times that of the sun. The surface temperature of sun is 6000 K. Calculate the temperature of the star.
Ans: E1 / E2 = (T1 / T2)^4
=> (1 / 17000) = (6000 / T2)^4
=> (1 / 17000)^1/4 = 6000 / T2
=> 1 / 11.418 = 6000 / T2
=> T2 = 68511 K
Que-23: Two stars, A and B, emit maximum radiations at wavelengths 5200 Å and 6500 Å respectively. If the temperature of A is 6000 K, then what is the temperature of B?
Ans: λ1 / λ2 = T2 / T1
=> 5200 / 6500 = T2 / 6000
=> 4 / 5 = T2 / 6000
=> T2 = 4800 K
Que-24: The sun and the moon emit maximum radiations at 5000 Å and 15 μ wavelengths, respectively. If the temperature of the sun be 6000 K, calculate the temperature of the moon.
Ans: λ1 / λ2 = T2 / T1
=> (5000 x 10^-10) / (15 x 10^-6) = T2 / 6000
=> 0.5 / 15 = T2 / 6000
=> T2 = 6000 x 0.5 / 15 = 200 K
Que-25: In solar-radiation the value of λm is 4753 Å and the temperature of the sun is 6080.3 K. Calculate the Wein’s constant b.
Ans: λ m T = b
=> 4753 x 10^-10 x 6080.3
=> 2.89 x 10^-3 m K
— : end of Heat Transfer Numerical on Stefan’s Law and Wien’s Law Class-11 Nootan ISC Physics Solutions :–
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