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Nookie1986 [14]

2 years ago

10

Sadi Carnot came up with a hypothetical heat engine that had the maximum possible efficiency. What discovery did Carnot make tha

t helped him recognize how to create this "perfect" heat engine?

1 answer:

otez555 [7]2 years ago

3 0

The discovery which Carnot made was that THE DIFFERENCE IN THE TEMPERATURES BETWEEN THE HOT AND THE COLD RESERVOIRS DETERMINE HOW WELL A HEAT ENGINE WOULD WORK.
Sadi Carnot was a French engineer, He proposed a theoretical thermodynamic cycle in 1824. In his cycle, Said hold that the efficiency of a heat engine depends on the temperature difference between its hot reservoir and cold reservoir.

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Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

lana [24]

The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
</span><span>
Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

7 0

3 years ago

Read 2 more answers

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

1 year ago

The temperatureTdin degrees Fahrenheit in terms of the Celsius temperaturedis given by=Td+9/5d+32.The temperatureCvin degrees Ce

Shalnov [3]

Answer:

$T_F=\frac{9}{5}(T_K-273)+32$

Explanation:

Since the temperature in degrees Fahrenheit in terms of the Celsius is given by the formula $T_F=\frac{9}{5}T_C+32$ and the temperature in degrees Celsius in terms of the Kelvin temperature is given by the formula $T_C=T_K-273$ , we can use the second formula and substitute it straight into the first formula (since a simplification is not being asked), obtaining $T_F=\frac{9}{5}(T_K-273)+32$ .

7 0

3 years ago

A sample of a diatomic ideal gas has pressure P and volume V. When the gas is warmed, its pressure triples and its volume double

Andreyy89

Answer:

Amount of Energy transferred $=8.5PV$

Explanation:

Given:

Initial volume=V

Initial pressure=P

Final volume=2V

Final pressure=3P

Now w know that the Energy transferred in constant pressure pressure is given by

$E_1=nc_pdT\\\\E_1=n\times\dfrac{7R}{2}dT\\E_1=3.5(nRdT)\\E_1=3.5(V(2P-P))\\E_1=3.5PV$

Now the Energy transferred in constant volume process is given by

$E_2=nc_vdT\\\\E_2=n\times\dfrac{5R}{2}dT\\E_1=2.5(nRdT)\\E_1=2.5(V(3V-V))\\E_1=5PV$

The total Energy transferred is given by

$E_{total}=E_1+E_2\\E{total}=3.5PV+5PV\\=8.5PV$

3 0

3 years ago

Read 2 more answers

Help me please I have other ones like this too on my page please help!

Dvinal [7]

Balanced equation is:
2Mn+4CuCl——->4Cu+2MnCl2
a=2
b=4
c=4
d=2

5 0

2 years ago