Ask Question
22 June, 01:32

An engine manufacturer makes the claim that the engine they have developed will, on each cycle take 100 J of heat out of boiling water at 100 °C, do mechanical work of 80J, and exhaust 20 J of heat at 10 °C.

What, if anything, is wrong with this claim?

A) There is nothing wrong with this claim because 100 J = 20 J + 80 J.

B) This engine violates the first law of thermodynamics because 100 J + 20 J ≠ 80 J.

C) An engine would operate by taking in heat at the lower temperature and exhausting heat at a higher temperature.

D) The heat exhausted must always be greater than the work done according to the second law of thermodynamics.

E) The efficiency of this engine is greater than the ideal Carnot cycle efficiency.

+1
Answers (1)
  1. 22 June, 03:42
    0
    Heat absorbed, QH = 100 J

    Heat exhausted, Qc = 20 J

    Mechanical work, W = 80 J

    Temperature of hot surface, TH = 100°C = 373 K

    Temperature of cold surface, Tc = 10°C = 283 K

    According to the carnot's engine

    TH / Tc = 373 / 283 = 1.32

    And QH / Qc = 100 / 20 = 5

    They are not equal to each other, so the engine is not possible.

    Option (D)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “An engine manufacturer makes the claim that the engine they have developed will, on each cycle take 100 J of heat out of boiling water at ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers