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timama [110]
4 years ago
12

Why must the operating temperature of a heat engine be higher than that of the cold sink?

Physics
1 answer:
Anna35 [415]4 years ago
6 0

Answer:

It is important to note, that the 2nd Law of thermodynamics plays no fundamental role in answering this question; we need a heat sink because the entropy is a state function, and at the end of the reversible process (which is visualized through the Carnot cycle diagram relevant for this problem), the entropy value of the system must return to the value it had originally.

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Which part of the electromagnetic spectrum has a lower frequency than visible light?
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radio waves have lower frequency than visible light

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g An insulated piston–cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred
Andrews [41]

The question is not complete and the complete question is;

An insulated piston-cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-Vdiagram with respect to saturation lines.

Answer:

Voltage of source = 175.33 V

I've attached an image of the P-Vdiagram of the process with respect to saturation lines.

Explanation:

To solve this question, we'll adopt the following assumptions.

- The kinetic and potential energy changes are negligible

-The thermal energy stored in the cylinder is negligible

-The cylinder is well insulated and thus heat transfer is negligible.

If the contents of the cylinder is taken as the system, the energy balance is;

Ein - Eout = ΔEsystem

Thus, [W(e-internal) + W(pw-internal)] - Wout = ΔU

Thus, IVΔt + W(pw-internal) = ΔH

= m(h2 - +1)

Now,looking at steam table A-5 attached to this answer, at P =175, x1 is calculated to be 0; dryness fraction, x2 = 0.5; vf = 0.001057m³/kg and hf = 487.01KJ/Kg and hg at saturation vapour = 2700.2

So, from this question v1=vf = 0.001057m³/kg and h1 = hf = 487.01KJ/Kg.

Also, h2 = hf + x2 (hg - hf) = 487.1 + 0.5(2700. 2 - 487.1) = 1593.65 KJ/Kg

The mass of the water is defined as;

M=V/v1

Volume(V) = 4L from the question. Since v1 is in m³/kg, let's convert V to m³. So V=4/1000 = 0.004

So, M = 0.004/0.001057 = 3.784kg

Now the formula for the voltage is ;

V = [M(h2 - h1) - W(pw-internal)] /(IΔt)

Δt=45minutes. We convert it to seconds to get 45x60 =2700 seconds

V = [[3.784 x (1593.65 - 487.1) - 400]/(8 x2700)] x 1000 = 175.33 V

6 0
3 years ago
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