A Carnot engine operates between 400 K and 300 K. Its efficiency is:
A. 33.3%
B. 20%
C. 25%
D. 75%
Answer: Option C
Solution (By JKSSB Mock Tests)
Carnot efficiency η = 1 - T₂/T₁, where T₁ is source temperature, T₂ sink temperature (in Kelvin). Here T₁=400 K, T₂=300 K, so η = 1 - 300/400 = 1 - 0.75 = 0.25 = 25%. This maximum possible efficiency for given temperatures is a fundamental thermodynamics result. Memory tip: 'η_Carnot = 1 - T_cold/T_hot'. Competitive exams frequently test this formula with varying temperatures. Always use absolute temperatures (Kelvin); Celsius would yield incorrect efficiency. This problem assesses understanding of heat engine limitations.
Explanation:
1 kWh = 1000 W × 3600 s = 3,600,000 J = 3.6 × 10⁶ J. This is a standard conversion. Power (watt) × time (hour) gives energy. 1 watt = 1 J/s. Remember this for electricity bills.
Explanation:
In the visible spectrum (VIBGYOR), wavelength increases from Violet to Red. Since speed = frequency × wavelength (c = fλ), frequency is inversely proportional to wavelength. Red light has the longest wavelength and therefore the lowest frequency and lowest energy. Violet has the shortest wavelength and highest frequency.
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