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ipn [44]
3 years ago
7

Air is contained inside a vertical piston-cylinder assembly by a piston of mass 105 kg and face area of 0.05 m2 . The mass of ai

r inside the cylinder is 20 grams, and the air initially occupies a volume of 50 liters. The atmosphere exerts a pressure of 101.3 kPa on the top of the piston. The volume of the air slowly decreases to 0.005 m3 , while maintaining constant pressure, as the specific internal energy of the air decreases by 175 kJ/kg. Neglecting friction between the piston and the cylinder wall, determine the heat between the surroundings and the air inside the piston-cylinder assembly in kJ
Engineering
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

ΔQ = 1.06 KJ

Explanation:

The amount of heat transfer between the piston-cylinder system and the surrounding can easily be found by using the First Law of Thermodynamics. The first law of thermodynamics can be written as follows:

ΔQ = ΔU + W

ΔQ = mΔu + PΔV

where,

ΔQ = Heat transfer between system and surrounding = ?

Δu = specific change in internal energy of the system = - 175 KJ/kg

m = mass of air = 20 g = 0.02 kg

P = Constant Pressure = 101.3 KPa

ΔV = Change in Volume = 0.05 m³ - 0.005 m³ = 0.045 m³

Therefore,

ΔQ = (0.02 kg)(-175 KJ/kg) + (101.3 KPa)(0.045 m³)

ΔQ = -3.5 KJ + 4.56 KJ

<u>ΔQ = 1.06 KJ</u>

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Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 gr
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(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.

<h3>Weight distribution of the kitten</h3>

In a normal distribution curve;

  • 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
  • 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
  • 1 standard deviation (d) above the mean (M), (M + d) is at 84%
  • 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

M - 2d = 125 g - 2(15g) = 95 g

M - d = 125 g - 15 g = 110 g

95 g is at 2% and 110 g is at 16%

(16% - 2%) = 14%

(110 - 95) = 15 g

14% / 15g = 0.93%/g

From 95 g to 99 g:

99 g - 95 g  = 4 g

4g x 0.93%/g = 3.72%

99 g will be at:

(2% + 3.72%) = 5.72%

Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

<h3>Weight of the kitten in the 90th percentile</h3>

M + d = 125 + 15 = 140 g      (at 84%)

M + 2d = 125 + 2(15) = 155 g   ( at 98%)

155 g - 140 g = 15 g

14% / 15g = 0.93%/g

84% + x(0.93%/g) = 90%

84 + 0.93x = 90

0.93x = 6

x = 6.45 g

weight of a kitten in 90th percentile = 140 g + 6.45 g  = 146.45 g

Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

Learn more about standard deviation here: brainly.com/question/475676

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