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

N76 [4]

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.

7 0

3 years ago

2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting po

Sonja [21]

Answer:

a) t = 11.2 s

b) v = 70.5 mph

Explanation:

a)

Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:

$t = \frac{v_{f} - v_{o}}{a} (1)$

where vf = 50 mph, and v₀ = 10 mph.
However, we still lack the value of a.
Assuming that the acceleration is constant, we can use the following kinematic equation:

$v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)$

Since we know that Δx = 500 ft, we could solve (2) for a.
In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:

$v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)$

$v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)$

We can do the same process with Δx, from ft to m, as follows:

$\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)$

Replacing (3), (4), and (5) in (2) and solving for a, we get:

$a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)$

Replacing (6) in (1) we finally get the value of the time t:

$t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)$

b)

Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:

$v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)$