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marissa [1.9K]
3 years ago
6

Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.

Physics
1 answer:
N76 [4]3 years ago
7 0

Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

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The heater element of a particular 120-V toaster is a 8.9-m length of nichrome wire, whose diameter is 0.86 mm. The resistivity
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Answer:

Power, P = 722.96 watts

Explanation:

It is given that,

Voltage, V = 120 V

Length of nichrome wire, l = 8.9 m

Diameter of wire, d = 0.86 mm

Radius of wire, r = 0.43 mm = 0.00043 m

Resistivity of wire, \rho=1.3\times 10^{-6}\ \Omega-m

We need to find the power drawn by this heater. Power is given by :

P=\dfrac{V^2}{R}

And, R=\rho\dfrac{l}{A}

P=\dfrac{V^2\times A}{\rho\times l}

P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}

P = 722.96 watts

So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.    

3 0
3 years ago
What element has 81 protons in the nuclei of atoms
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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo
makkiz [27]

Answer:

(a) The speed of the target proton after the collision is:V_{2f} =433(m/s), and (b) the speed of the projectile proton after the collision is: v_{1f}=250(m/s).

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}, and y axle:0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}. Now replacing the value given as: v_{1i}=500(m/s), \beta_{1}=+60^{o} for the projectile proton and according to the problem \beta_{1}and\beta_{2} are perpendicular so \beta_{2}=-30^{o}, and assuming that m_{1}=m_{2}, we get for x axle:500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2} and y axle: 0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}, then solving for v_{2f}, we get:v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f} and replacing at the first equation we get:500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}, now solving for v_{1f}, we can find the speed of the projectile proton after the collision as:v_{1f}=250(m/s) and v_{2f}=\sqrt{3}*v_{1f}=433(m/s), that is the speed of the target proton after the collision.

5 0
2 years ago
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Resistance = 2(1 kΩ) = 2 kΩ
8 0
3 years ago
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