Heat Transfer and Solar Energy
NIOS · Class 12 · Physics
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Which mode of heat transfer does NOT require a material medium for propagation?
The SI unit of thermal conductivity (K) is:
Among the following materials, which has the HIGHEST thermal conductivity?
Wien's displacement law states that λ_m × T = constant. If the surface temperature of the Sun is 6000 K and the Wien's constant is 2.88 × 10⁻³ m·K, what is the peak wavelength of radiation emitted by the Sun?
Sample Questions
According to Stefan-Boltzmann law, the energy radiated per second by a surface is proportional to which power of its absolute temperature?
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Fourth power (T⁴)
Step 1: The Stefan-Boltzmann law states: E = e·σ·A·T⁴. Step 2: Here, E is the energy radiated per second (power in watts), σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ is the Stefan-Boltzmann constant. Step 3: The temperature T must be in Kelvin (not Celsius), and the relationship is with T⁴. Step 4: This means if temperature doubles (T → 2T), the energy radiated increases by a factor of 2⁴ = 16 times. Step 5: Newton's Law of Cooling (E ∝ T – T₀) is a linear relationship and is only a special approximation of the Stefan-Boltzmann law for small temperature differences.
Which gas in the atmosphere is primarily responsible for the greenhouse effect by trapping infrared radiation?
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Carbon dioxide (CO₂)
Step 1: The greenhouse effect is the trapping of heat (infrared radiation) in the Earth's atmosphere. Step 2: The atmosphere is transparent to visible (short-wavelength) light from the Sun – it passes through and warms the Earth's surface. Step 3: The Earth's surface then re-radiates this energy as infrared (long-wavelength) radiation. Step 4: CO₂ in the atmosphere is opaque to infrared radiation – it reflects it back to Earth, raising the surface temperature. Step 5: Oxygen and Nitrogen make up ~99% of the atmosphere but are transparent to infrared. CO₂, though present in trace amounts, has a
The solar constant for Earth is approximately:
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1.36 × 10³ W m⁻²
Step 1: The solar constant is defined as the amount of solar energy received per unit area per second at the Earth's surface (perpendicular to sunlight). Step 2: Its experimentally measured value is 1.36 × 10³ W m⁻² = 1360 W m⁻² (approximately 1.36 kW m⁻²). Step 3: This is used to calculate the total power received by Earth: Q = 2πR²C, where R is Earth's radius. Step 4: Q = 2 × 3.14 × (6.4 × 10⁶)² × 1360 ≈ 3.5 × 10¹⁷ W. Step 5: Note: Only half of Earth's surface is illuminated at any time, hence 2πR² (hemisphere) is used.
For a perfectly black body, the value of absorptive power (a) is:
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1
Step 1: A perfectly black body is defined as one that absorbs ALL incident radiation of every wavelength. Step 2: The absorptive power 'a' is the fraction of incident energy absorbed: a = Energy absorbed / Energy incident. Step 3: Since a perfectly black body absorbs 100% of incident energy, a = 1. Step 4: For a perfectly black body: reflectivity (r) = 0, transmissivity (t) = 0, and a = 1, satisfying r + a + t = 1. Step 5: A perfectly black body is also the best emitter of radiation at any given temperature. Lamp black (absorbs ~96%) and platinum black (~98%) are close approximations in practi
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