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BaLLatris [955]
3 years ago
8

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

e mixture at constant pressure until the volume is doubled. Assuming constant specific heats at the average temperature, determine (a) the heat transfer and (b) the entropy change of the mixture. 13–68E During the expansion process of the ideal Otto
Engineering
1 answer:
Taya2010 [7]3 years ago
5 0

Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

 cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

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Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca
Andru [333]

Answer:

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

Explanation:

<u>a)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in series  

<u>List the Knowns: </u>

Capacitance of the first capacitor: C_{1}= 2цF = 2 x 10-6 F

Capacitance of the second capacitor C_{2}= 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

<u>Set Up the Problem:   </u>

Capacitance of a series combination:  

\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }+............

\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\

Capacitance of a series combination is given by:

C_{s}=\frac{Q}{V}

Then the charge stored in the series combination is:  

Q=C_{s} V

Energy stored in the series combination is:  

U_{c}=\frac{1}{2}  V^{2} C_{s}

<u>Solve the Problem:  </u>

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

<u>b)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in parallel  

<u>Set Up the Problem:  </u>

Capacitance of a parallel combination:

C_{p} =C_{1} +C_{2} +C_{3}

C_{p} =2+7.4=9.4*10^-6F

Capacitance of a parallel combination is given by

C_{p} =\frac{Q}{V}

Then the charge stored in the parallel combination is

Q=C_{p} V

Energy stored in the parallel combination is:  

U_{c}=\frac{1}{2} V^2C_{p}

<u>Solve the Problem: </u><em>  </em>

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

5 0
2 years ago
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To make 1000 containers of ice cream you need: 600 gallons of milk, 275 gallons of cream, and 120 gallons of flavor. Each ingred
Kamila [148]

Answer:

For the cream, 32 gallons should be reduced and 12 gallons should be decreased for flavor.

Explanation:

To prepare a total of 1000 gallons of ice cream you need 600 gallons of milk, 275 gallons of cream and 120 gallons of flavor, therefore we must calculate the percentages of each ingredient, as follows:

%milk=(600/1000)x100=60%

%cream=(275/1000)x100=27.5%

%flavor=(120/1000)x100=12%

If you reduce the amount of milk by 10% you have:

Milk quantity=600 gallons-(600x0.1)=540 gallons

To maintain the same percentages of each ingredient, you must make a rule of three to know the amount of cream and flavor that would need to be used with the 540 gallons of milk. The rule of three is as follows:

540 gallons of milk------------------60%

x gallons of cream--------------------27.5%

Clearing the x:

x gallons of cream=(540x27.5)/60=243 gallons

In the same way for flavor:

540 gallons of milk------------------60%

x gallons of flavor--------------------12%

Clearing the x:

x gallons of flavor=(540x12)/60=108 gallons

Verifying that they meet the percentages that were calculated before:

Total amount of ice cream=540+243+108=891 gallons

Calculate the percentages of each ingredient:

%milk=(540/891)x100=60.6%

%cream=(243/891)x100=27.3%

%flavor=(108/891)x100=12.1%

As can be seen, it is found that approximately the same percentages calculated above are met. Therefore, we can already calculate the amount by which the cream should be reduced and the flavor.

For the cream:

Gallons of cream=275-243=32 gallons

For the flavor:

Gallons of flavor=120-108=12 gallons

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Select the properties and typical applications for the high carbon steels.
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Answer:

<u>Option-(A)</u>

Explanation:

<u>Typical applications for the high carbon steels includes the following;</u>

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

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irina1246 [14]

Answer:A. 40% B.50% C. 60% Od 70%

Explanation:A. True B. False

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