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Lina20 [59]
3 years ago
6

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

11 °C. Determine the total heat transfer to this system in kJ. Give your answer with three decimal points
Engineering
1 answer:
labwork [276]3 years ago
7 0

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

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4.modifying a building to achieve an entirely new purpose

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Select the creative imaging fields that require knowledge of programming.
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Video game designer for sure
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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636
luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

<u>L1 = 76.093 mm</u>

5 0
3 years ago
Write the definition of a function printLarger, which has two int parameters and returns nothing. The function prints the larger
vredina [299]

Answer:

  1. #include <iostream>
  2. using namespace std;
  3. void printLarger(int a, int b){
  4.    
  5.    if(a > b){
  6.        cout<<a;
  7.    }else{
  8.        cout<<b;
  9.    }
  10. }
  11. int main()
  12. {
  13.    printLarger(4, 5);
  14.    return 0;
  15. }

Explanation:

The solution code is written in C++.

Firstly define a function printLarger that has two parameters, a and b with both of them are integer type (Line 5). In the function, create an if condition to check if a bigger than b, print a to terminal (Line 7-8). Otherwise print b (Line 9-10).

In the main program, test the function by passing 4 and 5 as arguments (Line 16) and we shall get 5 printed.  

7 0
3 years ago
A corn ethanol production plant receives 500,000.0 kg/day corn feedstock at a moisture content of 15.5% (wet basis). If all of t
Alenkasestr [34]

Answer:

207 m³/day

Explanation:

Dry corn feed stock = 500000 × ( 100 - 15.5%) = 500000 × 84.5% = 500000× 0.845 = 422500

Starch yield = 68.5% × 422500 = 289412.5

Glucose yield = 1.11 × 289412.5 = 321247.875 where 1.11 is the scarification factor of starch to glucose

Ethanol yield = 0.51 × 321247.875 = 163836.416 where 0.51 is theoretical yield of ethanol from one mole of glucose

density = mass / volume

volume = mass / density = 163836.416 / 789 = 207 m³ / day

4 0
3 years ago
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