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AleksandrR [38]
3 years ago
15

What happens to the particles in water as the water is heated and turns to vapor

Physics
2 answers:
lions [1.4K]3 years ago
6 0
Hey There!

Your answer is that the particles will move more faster!

In a IceCube the particles are badly moving in a liquid they move more but not fast and in this case as the particles are heating changing into a gas the particles are moving faster!

If you need anymore help just ask me!

Hope this helps!
vichka [17]3 years ago
6 0

Answer: The particles move more quickly.

Explanation:

Kinetic energy is the energy possessed by an object by virtue of its motion. It is determined by the equation:

K=\frac{3RT}{2}

From above, it is visible that kinetic energy is directly related to the temperature of the system. So, if temperature is more, average kinetic energy of the system is more and movement of particles is more.

Thus the particles in water are heated, they gain kinetic energy and hence move more quickly.

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What amount of heat is required to raise the temperature of 200 g of water by 15°C (the specific heat of water is 1 cal/g°C)
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Answer:

Heat energy required (Q) = 3,000 J

Explanation:

Find:

Mass of water (M) = 200 g

Change in temperature (ΔT) = 15°C

Specific heat of water (C) = 1 cal/g°C

Find:

Heat energy required (Q) = ?

Computation:

Q = M × ΔT × C

Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)

Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C

Heat energy required (Q) = 3,000 J

4 0
3 years ago
1 point
klio [65]

Answer:

0

10

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3 years ago
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

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3 years ago
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