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3 years ago

Which statement best describes how a wave would move differently through a pot of boiling water than the steam created from it?

Physics

2 answers:

Marina86 [1]3 years ago

3 0

<span>The wave would move faster through the water than through the steam.</span>

shepuryov [24]3 years ago

3 0

Hello. This question is incomplete. The complete question is:

"Which statement best describes how a wave would move differently through a pot of boiling water than the steam created from it? The wave would move faster through the water than through the steam. The wave would move slower through the water than through the steam. The wave would move the same speed through the water and the steam. The wave would move through the water and steam at the same speed, but decrease in speed at the transition point. "

Answer:

The wave would move faster through the water than through the steam.

Explanation:

As we already know, the water molecules at the moment when the water is boiling move very fast, due to the thermal energy that acts on them. This movement causes the mileculas to collide with each other quickly, for this reason we can observe that movement of the boiling water. This makes the wave movement in the boiling water fast.

On the other hand, all this movement is not seen in water vapor molecules, since the space between them is larger and there is no energy acting directly on them. For this reason, we can say that, through steam, the wave would not move with such speed.

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An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of

Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$ (taking the value of $v_g$ and $v_g$ at 200°C )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

$=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

$=850.65+3.377\times10^{-3}\times 1744.65$

= 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

= 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

= 1.51 kW

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