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SCORPION-xisa [38]
2 years ago
13

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

ept at so that the engine efficiency is 0.79?
Physics
1 answer:
Paraphin [41]2 years ago
6 0
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
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It's mostly <em>convection</em> ... hot air and steam rising from the soup to your hand.

Then, of course, there HAS to be some conduction when the hot gases reach your hand ... their heat has to soak into your skin, and that's conduction.

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1 year ago
Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a
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E=0

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4 0
1 year ago
The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o
MrMuchimi

Answer:

The total kinetic energy of both particles is 2.43\times10^{-13}

Explanation:

Given that,

Kinetic energy of nucleusK.E= 1.24\times10^{-13}\ J

Kinetic energy of proton K.E= 2.47\times10^{-13}\ J

Radius of proton r= 0.9\times10^{-15}\ m

We need to calculate the final potential energy

Using formula of final potential energy

U=\dfrac{kq^2}{r}

Put the value into the formula

U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}

U_{f}=1.28\times10^{-13}\ J

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Using formula of energy

E_{i}=(K.E_{n}+K.E_{p})+U_{i}

E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0

E_{i}=3.71\times10^{-13}\ J

We need to calculate the total kinetic energy of both particles

Using conservation of energy

E_{i}=E_{f}

E_{i}=K.E_{f}+U_{f}

3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}

K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}

K.E_{f}=2.43\times10^{-13}

Hence, The total kinetic energy of both particles is 2.43\times10^{-13}

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V = - 1.06 m/s

Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.

5 0
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