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Bezzdna [24]
3 years ago
8

There are two piston-cylinder systems that each contain 1 kg of an idea gas at a pressure of 300 kPa and temperature of 350 K. T

he two systems then undergo two different processes:
System 1 undergoes process 1 as follows:
Isobaric heating to 500 K, then compressed to a pressure of 500 kPa and a temperature of 600 K.

System 2 undergoes process 2 as follows:
Directly compressed in one step to a pressure 500 kPa and a temperature of 600 K

We "measure" the entropy of each system before and after the process. Which system had the biggest entropy change?

a. System 1
b. System 2
c. Entropy change is the same for both systems
Engineering
1 answer:
Nataly_w [17]3 years ago
8 0

Answer:

Entropy change is same for both systems because entropy is a state function i.e., it doesn't depend on the path by which the system arrived at it's present state (500 kPa, 600 K) from initial state (300 kPa, 350 K) .

Explanation:

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Would be much appreciated if someone could help with this will give brainiest.
Mashcka [7]

Answer:   both mm and inches on each dimension in a sketch (with the main dimension in one format and the other in brackets below it), in the way you can have dual dimensions shown when detailing an idw view.

personally think it would look a mess/cluttered with even more text all over the sketch environment, but everyone's differenent.

If it's any help - you know you can enter dimensions in either format?  If you're working in mm you can still dimension a line and type "2in" and vice-versa.  Probably know this already, but no harm saying it, just in case.

You can enter the units directly in or mm and Inventor will convert to current document settings (which  you can change - maybe someone can come up with a simple toggle icon to toggle the document settings).  Tools>Document Settings>Units

Unlike SolidWorks when you edit the dimension the original entry shows in the dialog box so it makes it easy to keep track of different units even if they aren't always displayed.  (SWx does the conversion or equation and then that is what you get.)

I work quite a bit in inch and metric and combination (ex metric frame motor on inch machine) and it doesn't seem to be a real difficulty to me.

4 0
3 years ago
Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa
netineya [11]

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C  = 50 K

m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}

2. Energy in British thermal units

\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}

7 0
3 years ago
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miss Akunina [59]

Answer:

Under no circumstances

Explanation:

I'm not 100% sure why, but I remember hearing that you're not suposed to go over the speed limit no matter what

7 0
3 years ago
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Pachacha [2.7K]

Answer:

im kinda smart  whyy?

Explanation:

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Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

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