A heat engine absorbs 150 J of heat from a hot reservoir and rejects 90 J to a c

Question

A heat engine absorbs 150 J of heat from a hot reservoir and rejects 90 J to a cold reservoir. What is the efficiency of this engine? A) 20% B) 40% C) 60% D) 67% E) 90% The maximum theoretical thermal efficiency of a steam engine that is supplied steam at a temperature of 600 degree C and exhausts it at a temperature of 200 degree C is A) 33.3% B) 45.8% C) 66.7% D) 77.1% E) 84.6% The coefficient of linear expansion of the steel in a certain bridge is 1.10 times 10^-5/C degree. If at a minimum winter temperature of -30 degree C the bridge is 30.5 m long, at the summer maximum of 40 degree C its length is increased about A) 2.54 cm B) 2.54 mm C) 0.254 mm D) 25.4 mum E) 2.54 mum If the thickness of a uniform wall is doubled, the rate at which heat is conducted through the wall is A) doubled. B) increased by a factor of four. C) decreased by a factor of four. D) cut in half. E) unchanged. Two small spheres, each with mass m = 5.0 g and charge q, are suspended from a point by threads of length L = 0.30 m. What is the charge on each sphere if the threads make an angle theta = 20 degree with the vertical? A) 7.9 times 10^-7 C D) 6.3 times 10^13 C B) 2.9 times 10^-7 C E) 1.8 times 10^-7C C) 7.5 times 10^-2 C Point charges of 4.0 times 10^-8 C and -2.0 times 10^-8 C are placed 12 cm apart. A third point charge of 3.0 times 10^-8 C halfway between the first two point charges experiences a force of magnitude A) 4.5 times 10^-3 N B) 2.0 times 10^-3 N C) 1.5 times 10^-3 N D) zero E) 5.0 times 10^-3 N

Explanation / Answer

14. Qin = 150 J

W = 150 - 90 = 60 J

n = W /Qin = 60 / 150 = 0.4

% efficiency = 0.4 x 100 = 40 %

Ans(B)

15. n = 1 - [ (273 + 200) / (273 + 600)]

n = 0.458

% efficiency = 45.8 %

16. deltaL = L0 alpha deltaT

= 30.5 x 1.10 x 10^-5 x (40 + 30)

= 0.0254 m

= 2.54 cm

Ans(A)

17. Q/t = k A deltaT / d

d - > doubled.

Q/t -> halved

Ans(D)

18. T cos20 = m g

T = (5 x 10^-3 x 9.8) / cos20

T = 0.0521 N

T sin20 = k q^2 / r^2

0.0521 sin20 = 9 x 10^9 x q^2 / (2 x0.30 x sin20)^2

q = 2.9 x 10^-7 C

ans(B)

19. F = (9 x 10^9 )(6 x 10^-8) (3 x 10^-8) / (0.06^2)

F = 4.5 x 10^-3 N

Ans(A)

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