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r-ruslan [8.4K]
3 years ago
6

Room temperature water is placed in an Erlenmyer flask and heated to the boiling point. After the flask is removed from the heat

source boiling ceases. Then a rubber stopper is placed in the mouth of the Erlenmyer flask to seal it from outside air. The flask is then placed in an ice bath. After a few moments the water begins to boil again. What phenomenon causes the water to boil again?
Physics
1 answer:
Vaselesa [24]3 years ago
7 0

Explanation:

When water is boiled in the flask . Some portion of it is evaporated out . Now when cork is placed on it and is placed in the ice box . It cools down , by which the pressure inside decreases .

Due to decrease of pressure , the boiling point of water also decreases . Now it can boil at lower temperature . Thus it starts boiling at lower temperature even , when placed in the ice box .

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8 0
2 years ago
As 390 g of hot milk cools in a mug it transfers 30 000 j of heat to the environment. whats is the temperature change of the mil
Fed [463]
You have to use the specific heat equation. 

Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.

So we can substitute our variables into the equation.

30000J = (390g)(3.9J*g/C)ΔT

Solving for ΔT, we get:

30000J/[(390g)*(3.9J*g/C) = ΔT

ΔT = 19.72386588C

I'm assuming the temperature is C, since it was not specified.

Hope this helps!
8 0
3 years ago
Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in
gtnhenbr [62]

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

  • q = charge on capacitor and
  • v = voltage across capacitor.

<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

  • q = charge on capacitor and
  • v = voltage across capacitor.

So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

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3 0
2 years ago
4. Explain how states of matter change in regards to<br> a. Temperature-<br> b. Pressure-
Alina [70]
Temperature can change the state from solid to liquid causing it to melting, liquid to gas causing vaporization or a solid to a gas causing sublimation. Pressure alone cannot change the state of matter.
5 0
3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0
3 years ago
Read 2 more answers
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