A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.
1 answer:
Answer:
a)
b) Efficiency=76.77%
Explanation:
a)
In order to solve this problem, we can use the following formula:
the problem provides us with all the necessary information so we can directly use the formula:
b) In order to find the efficiency, we can use the following formula:
so we get:
Efficiency=76.77%
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Answer:
the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Explanation:
Given the data in the question;
wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m
Index of refraction; n = 1.35
Now, the thinnest thickness of the soap film can be determined from the following expression;
= ( λ / 4n )
so we simply substitute in our given values;
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= ( 463 × 10⁻⁹ m ) / 5.4
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= 8.574 × 10⁻⁸ m
= 85.74 × 10⁻⁹ m
= 85.74 nm
Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm