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Nataliya [291]
3 years ago
15

A heat engine takes in 840 kJ per cycle from a heat reservoir. Which is not a possible value of the engine's heat output per cyc

le?
Physics
1 answer:
zhannawk [14.2K]3 years ago
8 0
We have by the first law of thermodynamics tha energy is preserved, hence we cannot have over 840kJ per cycle. We have by the laws of thermodynamics (the 2nd one in specific) that the entropy of a system cannot increase. We cannot have an output of  840 kJ per cycle from a heat engine because then that would mean that the entropy would stay the same, while any heat engine increases it. Hence, any value \geq 840 kJ is acceptable.
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We are given that,

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Differentiating each side with respect to t, we get,

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\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}

Replacing the value of the velocity

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\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta

The value of cos \theta could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

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Answer:

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